Line UMK G. K. Muravin. Algebra and principles of mathematical analysis (10-11) (in-depth)

Line UMK G.K. Muravina, K.S. Muravina, O.V. Muravina. Algebra and principles of mathematical analysis (10-11) (basic)

How to teach solving trigonometric equations and inequalities: teaching methods

The mathematics course of the Russian Textbook Corporation, authored by Georgy Muravina and Olga Muravina, provides for a gradual transition to solving trigonometric equations and inequalities in the 10th grade, as well as continuing their study in the 11th grade. We present to your attention the stages of transition to the topic with excerpts from the textbook “Algebra and the beginning of mathematical analysis” (advanced level).

1. Sine and cosine of any angle (propaedeutic to the study of trigonometric equations)

Example assignment. Find approximately the angles whose cosines are equal to 0.8.

Solution. The cosine is the abscissa of the corresponding point on the unit circle. All points with abscissas equal to 0.8 belong to a straight line parallel to the ordinate axis and passing through the point C(0.8; 0). This line intersects the unit circle at two points: P α ° And P β ° , symmetrical about the abscissa axis.

Using a protractor we find that the angle α° approximately equal to 37°. So, the general view of the rotation angles with the end point P α°:

α° ≈ 37° + 360° n, Where n- any integer.

Due to symmetry about the abscissa axis, the point P β ° - end point of rotation at an angle of –37°. This means that for her the general form of rotation angles is:

β° ≈ –37° + 360° n, Where n- any integer.

Answer: 37° + 360° n, –37° + 360° n, Where n- any integer.

Example assignment. Find the angles whose sines are equal to 0.5.

Solution. The sine is the ordinate of the corresponding point on the unit circle. All points with ordinates equal to 0.5 belong to a straight line parallel to the abscissa axis and passing through the point D(0; 0,5).

This line intersects the unit circle at two points: Pφ and Pπ–φ, symmetrical about the ordinate axis. In a right triangle OKPφ leg KPφ is equal to half the hypotenuse OPφ , Means,

General view of rotation angles with end point P φ :

Where n- any integer. General view of rotation angles with end point P π–φ :

Where n- any integer.

Answer: Where n- any integer.

2. Tangent and cotangent of any angle (propaedeutics for the study of trigonometric equations)

Example 2.

Example assignment. Find the general form of angles whose tangent is –1.2.

Solution. Let us mark the point on the tangent axis C with an ordinate equal to –1.2, and draw a straight line O.C.. Straight O.C. intersects the unit circle at points P α ° And Pβ° - ends of the same diameter. The angles corresponding to these points differ from each other by an integer number of half-turns, i.e. 180° n (n- integer). Using a protractor we find that the angle P α° OP 0 is equal to –50°. This means that the general form of angles whose tangent is –1.2 is as follows: –50° + 180° n (n- integer)

Answer:–50° + 180° n, n∈ Z.

Using the sine and cosine of angles of 30°, 45° and 60°, it is easy to find their tangents and cotangents. For example,

The listed angles are quite common in various problems, so it is useful to remember the values ​​of the tangent and cotangent of these angles.

3. The simplest trigonometric equations

The following notations are introduced: arcsin α, arccos α, arctg α, arcctg α. It is not recommended to rush into introducing the combined formula. It is much more convenient to record two series of roots, especially when you need to select roots at intervals.

When studying the topic “the simplest trigonometric equations,” the equations are most often reduced to squares.

4. Reduction formulas

Reduction formulas are identities, i.e. they are true for any valid values φ . Analyzing the resulting table, you can see that:

1) the sign on the right side of the formula coincides with the sign of the reducible function in the corresponding quadrant, if we consider φ acute angle;

2) the name is changed only by the functions of the angles and

	φ + 2π n

5. Properties and graph of a function y= sin x

The simplest trigonometric inequalities can be solved either on a graph or on a circle. When solving a trigonometric inequality on a circle, it is important not to confuse which point to indicate first.

6. Properties and graph of a function y=cos x

The task of constructing a graph of a function y=cos x can be reduced to plotting the function y= sin x. Indeed, since graph of a function y=cos x can be obtained from the graph of the function y= sin x shifting the latter along the x-axis to the left by

7. Properties and graphs of functions y= tg x And y=ctg x

Function Domain y= tg x includes all numbers except numbers of the form where n ∈ Z. As when constructing a sinusoid, first we will try to obtain a graph of the function y = tg x in between

At the left end of this interval, the tangent is zero, and when approaching the right end, the tangent values ​​increase without limit. Graphically it looks like the graph of a function y = tg x presses against the straight line, going upward with it unlimitedly.

8. Dependencies between trigonometric functions of the same argument

Equality and express relations between trigonometric functions of the same argument φ. With their help, knowing the sine and cosine of a certain angle, you can find its tangent and cotangent. From these equalities it is easy to see that tangent and cotangent are related to each other by the following equality.

tg φ · cot φ = 1

There are other dependencies between trigonometric functions.

Equation of the unit circle centered at the origin x 2 + y 2= 1 connects the abscissa and ordinate of any point on this circle.

Fundamental trigonometric identity

cos 2 φ + sin 2 φ = 1

9. Sine and cosine of the sum and difference of two angles

Cosine sum formula

cos (α + β) = cos α cos β – sin α sin β

Difference cosine formula

cos (α – β) = cos α cos β + sin α sin β

Sine difference formula

sin (α – β) = sin α cos β – cos α sin β

Sine sum formula

sin (α + β) = sin α cos β + cos α sin β

10. Tangent of the sum and tangent of the difference of two angles

Tangent sum formula

Tangent difference formula

The textbook is included in the teaching materials in mathematics for grades 10–11 studying the subject at a basic level. Theoretical material is divided into mandatory and optional, the system of tasks is differentiated by level of difficulty, each chapter ends with test questions and assignments, and each chapter with a home test. The textbook includes project topics and links to Internet resources.

11. Trigonometric double angle functions

Double angle tangent formula

cos2α = 1 – 2sin 2 α cos2α = 2cos 2 α – 1

Example assignment. Solve the equation

Solution.

13. Solving trigonometric equations

In most cases, the original equation is reduced to simple trigonometric equations during the solution process. However, there is no single solution method for trigonometric equations. In each specific case, success depends on knowledge of trigonometric formulas and the ability to choose the right ones from them. However, the abundance of different formulas sometimes makes this choice quite difficult.

Equations that reduce to squares

Example assignment. Solve equation 2 cos 2 x+ 3 sin x = 0

Solution. Using the basic trigonometric identity, this equation can be reduced to a quadratic equation with respect to sin x:

2cos 2 x+3sin x= 0, 2(1 – sin 2 x) + 3sin x = 0,

2 – 2sin 2 x+3sin x= 0, 2sin 2 x– 3sin x – 2 = 0

Let's introduce a new variable y= sin x, then the equation will take the form: 2 y 2 – 3y – 2 = 0.

The roots of this equation y 1 = 2, y 2 = –0,5.

Returning to the variable x and we get the simplest trigonometric equations:

1) sin x= 2 – this equation has no roots, since sin x < 2 при любом значении x;

2) sin x = –0,5,

Answer:

Homogeneous trigonometric equations

Example assignment. Solve the equation 2sin 2 x– 3sin x cos x– 5cos 2 x = 0.

Solution. Let's consider two cases:

1)cos x= 0 and 2) cos x ≠ 0.

Case 1. If cos x= 0, then the equation takes the form 2sin 2 x= 0, whence sin x= 0. But this equality does not satisfy the cos condition x= 0, since under no circumstances x Cosine and sine do not vanish at the same time.

Case 2. If cos x≠ 0, then we can divide the equation by cos 2 x “Algebra and the beginning of mathematical analysis. 10th grade”, like many other publications, is available on the LECTA platform. To do this, take advantage of the offer.

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The simplest trigonometric equations are the equations

Cos (x) = a, sin (x) = a, tg (x) = a, ctg (x) =a

Equation cos(x) = a

Explanation and rationale

1. The roots of the equation cosx = a. When | a | > 1 the equation has no roots, since | cosx |< 1 для любого x (прямая y = а при а >1 or at a< -1 не пересекает график функцииy = cosx).

Let | a |< 1. Тогда прямая у = а пересекает график функции

y = cos x. On the interval, the function y = cos x decreases from 1 to -1. But a decreasing function takes each of its values ​​only at one point of its domain of definition, therefore the equation cos x = a has only one root on this interval, which, by definition of arccosine, is equal to: x 1 = arccos a (and for this root cos x = A).

Cosine is an even function, so on the interval [-n; 0] the equation cos x = and also has only one root - the number opposite x 1, that is

x 2 = -arccos a.

Thus, on the interval [-n; p] (length 2p) equation cos x = a with | a |< 1 имеет только корни x = ±arccos а.

The function y = cos x is periodic with a period of 2n, therefore all other roots differ from those found by 2n (n € Z). We obtain the following formula for the roots of the equation cos x = a when

x = ±arccos a + 2pp, n £ Z.

1. Special cases of solving the equation cosx = a.

It is useful to remember special notations for the roots of the equation cos x = a when

a = 0, a = -1, a = 1, which can be easily obtained using the unit circle as a reference.

Since the cosine is equal to the abscissa of the corresponding point of the unit circle, we obtain that cos x = 0 if and only if the corresponding point of the unit circle is point A or point B.

Similarly, cos x = 1 if and only if the corresponding point of the unit circle is point C, therefore,

x = 2πп, k € Z.

Also cos x = -1 if and only if the corresponding point of the unit circle is point D, thus x = n + 2n,

Equation sin(x) = a

Explanation and rationale

1. The roots of the equation sinx = a. When | a | > 1 the equation has no roots, since | sinx |< 1 для любого x (прямая y = а на рисунке при а >1 or at a< -1 не пересекает график функции y = sinx).

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An equality containing an unknown under the sign of a trigonometric function (`sin x, cos x, tan x` or `ctg x`) is called a trigonometric equation, and it is their formulas that we will consider further.

The simplest equations are `sin x=a, cos x=a, tg x=a, ctg x=a`, where `x` is the angle to be found, `a` is any number. Let us write down the root formulas for each of them.

1. Equation `sin x=a`.

For `|a|>1` it has no solutions.

When `|a| \leq 1` has an infinite number of solutions.

Root formula: `x=(-1)^n arcsin a + \pi n, n \in Z`

2. Equation `cos x=a`

For `|a|>1` - as in the case of sine, it has no solutions among real numbers.

When `|a| \leq 1` has an infinite number of solutions.

Root formula: `x=\pm arccos a + 2\pi n, n \in Z`

Special cases for sine and cosine in graphs.

3. Equation `tg x=a`

Has an infinite number of solutions for any values ​​of `a`.

Root formula: `x=arctg a + \pi n, n \in Z`

4. Equation `ctg x=a`

Also has an infinite number of solutions for any values ​​of `a`.

Root formula: `x=arcctg a + \pi n, n \in Z`

Formulas for the roots of trigonometric equations in the table

For sine:
For cosine:
For tangent and cotangent:
Formulas for solving equations containing inverse trigonometric functions:

Methods for solving trigonometric equations

Solving any trigonometric equation consists of two stages:

• with the help of transforming it to the simplest;
• solve the simplest equation obtained using the root formulas and tables written above.

Let's look at the main solution methods using examples.

Algebraic method.

This method involves replacing a variable and substituting it into an equality.

Example. Solve the equation: `2cos^2(x+\frac \pi 6)-3sin(\frac \pi 3 - x)+1=0`

`2cos^2(x+\frac \pi 6)-3cos(x+\frac \pi 6)+1=0`,

make a replacement: `cos(x+\frac \pi 6)=y`, then `2y^2-3y+1=0`,

we find the roots: `y_1=1, y_2=1/2`, from which two cases follow:

1. `cos(x+\frac \pi 6)=1`, `x+\frac \pi 6=2\pi n`, `x_1=-\frac \pi 6+2\pi n`.

2. `cos(x+\frac \pi 6)=1/2`, `x+\frac \pi 6=\pm arccos 1/2+2\pi n`, `x_2=\pm \frac \pi 3- \frac \pi 6+2\pi n`.

Answer: `x_1=-\frac \pi 6+2\pi n`, `x_2=\pm \frac \pi 3-\frac \pi 6+2\pi n`.

Factorization.

Example. Solve the equation: `sin x+cos x=1`.

Solution. Let's move all the terms of the equality to the left: `sin x+cos x-1=0`. Using , we transform and factorize the left-hand side:

`sin x — 2sin^2 x/2=0`,

`2sin x/2 cos x/2-2sin^2 x/2=0`,

`2sin x/2 (cos x/2-sin x/2)=0`,

1. `sin x/2 =0`, `x/2 =\pi n`, `x_1=2\pi n`.
2. `cos x/2-sin x/2=0`, `tg x/2=1`, `x/2=arctg 1+ \pi n`, `x/2=\pi/4+ \pi n` , `x_2=\pi/2+ 2\pi n`.

Answer: `x_1=2\pi n`, `x_2=\pi/2+ 2\pi n`.

Reduction to a homogeneous equation

First, you need to reduce this trigonometric equation to one of two forms:

`a sin x+b cos x=0` (homogeneous equation of the first degree) or `a sin^2 x + b sin x cos x +c cos^2 x=0` (homogeneous equation of the second degree).

Then divide both parts by `cos x \ne 0` - for the first case, and by `cos^2 x \ne 0` - for the second. We obtain equations for `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, which need to be solved using known methods.

Example. Solve the equation: `2 sin^2 x+sin x cos x - cos^2 x=1`.

Solution. Let's write the right side as `1=sin^2 x+cos^2 x`:

`2 sin^2 x+sin x cos x — cos^2 x=` `sin^2 x+cos^2 x`,

`2 sin^2 x+sin x cos x — cos^2 x -` ` sin^2 x — cos^2 x=0`

`sin^2 x+sin x cos x — 2 cos^2 x=0`.

This is a homogeneous trigonometric equation of the second degree, we divide its left and right sides by `cos^2 x \ne 0`, we get:

`\frac (sin^2 x)(cos^2 x)+\frac(sin x cos x)(cos^2 x) — \frac(2 cos^2 x)(cos^2 x)=0`

`tg^2 x+tg x — 2=0`. Let's introduce the replacement `tg x=t`, resulting in `t^2 + t - 2=0`. The roots of this equation are `t_1=-2` and `t_2=1`. Then:

1. `tg x=-2`, `x_1=arctg (-2)+\pi n`, `n \in Z`
2. `tg x=1`, `x=arctg 1+\pi n`, `x_2=\pi/4+\pi n`, ` n \in Z`.

Answer. `x_1=arctg (-2)+\pi n`, `n \in Z`, `x_2=\pi/4+\pi n`, `n \in Z`.

Moving to Half Angle

Example. Solve the equation: `11 sin x - 2 cos x = 10`.

Solution. Let's apply the double angle formulas, resulting in: `22 sin (x/2) cos (x/2) -` `2 cos^2 x/2 + 2 sin^2 x/2=` `10 sin^2 x/2 +10 cos^2 x/2`

`4 tg^2 x/2 — 11 tg x/2 +6=0`

Applying the algebraic method described above, we obtain:

1. `tg x/2=2`, `x_1=2 arctg 2+2\pi n`, `n \in Z`,
2. `tg x/2=3/4`, `x_2=arctg 3/4+2\pi n`, `n \in Z`.

Answer. `x_1=2 arctg 2+2\pi n, n \in Z`, `x_2=arctg 3/4+2\pi n`, `n \in Z`.

Introduction of auxiliary angle

In the trigonometric equation `a sin x + b cos x =c`, where a,b,c are coefficients and x is a variable, divide both sides by `sqrt (a^2+b^2)`:

`\frac a(sqrt (a^2+b^2)) sin x +` `\frac b(sqrt (a^2+b^2)) cos x =` `\frac c(sqrt (a^2) +b^2))`.

The coefficients on the left side have the properties of sine and cosine, namely the sum of their squares is equal to 1 and their modules are not greater than 1. Let us denote them as follows: `\frac a(sqrt (a^2+b^2))=cos \varphi` , ` \frac b(sqrt (a^2+b^2)) =sin \varphi`, `\frac c(sqrt (a^2+b^2))=C`, then:

`cos \varphi sin x + sin \varphi cos x =C`.

Let's take a closer look at the following example:

Example. Solve the equation: `3 sin x+4 cos x=2`.

Solution. Divide both sides of the equality by `sqrt (3^2+4^2)`, we get:

`\frac (3 sin x) (sqrt (3^2+4^2))+` `\frac(4 cos x)(sqrt (3^2+4^2))=` `\frac 2(sqrt (3^2+4^2))`

`3/5 sin x+4/5 cos x=2/5`.

Let's denote `3/5 = cos \varphi` , `4/5=sin \varphi`. Since `sin \varphi>0`, `cos \varphi>0`, then we take `\varphi=arcsin 4/5` as an auxiliary angle. Then we write our equality in the form:

`cos \varphi sin x+sin \varphi cos x=2/5`

Applying the formula for the sum of angles for the sine, we write our equality in the following form:

`sin (x+\varphi)=2/5`,

`x+\varphi=(-1)^n arcsin 2/5+ \pi n`, `n \in Z`,

`x=(-1)^n arcsin 2/5-` `arcsin 4/5+ \pi n`, `n \in Z`.

Answer. `x=(-1)^n arcsin 2/5-` `arcsin 4/5+ \pi n`, `n \in Z`.

Fractional rational trigonometric equations

These are equalities with fractions whose numerators and denominators contain trigonometric functions.

Example. Solve the equation. `\frac (sin x)(1+cos x)=1-cos x`.

Solution. Multiply and divide the right side of the equality by `(1+cos x)`. As a result we get:

`\frac (sin x)(1+cos x)=` `\frac ((1-cos x)(1+cos x))(1+cos x)`

`\frac (sin x)(1+cos x)=` `\frac (1-cos^2 x)(1+cos x)`

`\frac (sin x)(1+cos x)=` `\frac (sin^2 x)(1+cos x)`

`\frac (sin x)(1+cos x)-` `\frac (sin^2 x)(1+cos x)=0`

`\frac (sin x-sin^2 x)(1+cos x)=0`

Considering that the denominator cannot be equal to zero, we get `1+cos x \ne 0`, `cos x \ne -1`, ` x \ne \pi+2\pi n, n \in Z`.

Let's equate the numerator of the fraction to zero: `sin x-sin^2 x=0`, `sin x(1-sin x)=0`. Then `sin x=0` or `1-sin x=0`.

1. `sin x=0`, `x=\pi n`, `n \in Z`
2. `1-sin x=0`, `sin x=-1`, `x=\pi /2+2\pi n, n \in Z`.

Given that ` x \ne \pi+2\pi n, n \in Z`, the solutions are `x=2\pi n, n \in Z` and `x=\pi /2+2\pi n` , `n \in Z`.

Answer. `x=2\pi n`, `n \in Z`, `x=\pi /2+2\pi n`, `n \in Z`.

Trigonometry, and trigonometric equations in particular, are used in almost all areas of geometry, physics, and engineering. Studying begins in the 10th grade, there are always tasks for the Unified State Exam, so try to remember all the formulas of trigonometric equations - they will definitely be useful to you!

However, you don’t even need to memorize them, the main thing is to understand the essence and be able to derive it. It's not as difficult as it seems. See for yourself by watching the video.

In this lesson we will look at basic trigonometric functions, their properties and graphs, and also list basic types of trigonometric equations and systems. In addition, we indicate general solutions of the simplest trigonometric equations and their special cases.

This lesson will help you prepare for one of the types of tasks B5 and C1.

Preparation for the Unified State Exam in mathematics

Experiment

Lesson 10. Trigonometric functions. Trigonometric equations and their systems.

Theory

Lesson summary

We have already used the term “trigonometric function” many times. Back in the first lesson of this topic, we defined them using a right triangle and a unit trigonometric circle. Using these methods of specifying trigonometric functions, we can already conclude that for them one value of the argument (or angle) corresponds to exactly one value of the function, i.e. we have the right to call sine, cosine, tangent and cotangent functions.

In this lesson, it's time to try to abstract from the previously discussed methods of calculating the values ​​of trigonometric functions. Today we will move on to the usual algebraic approach to working with functions, we will look at their properties and depict graphs.

Regarding the properties of trigonometric functions, special attention should be paid to:

The domain of definition and the range of values, because for sine and cosine there are restrictions on the range of values, and for tangent and cotangent there are restrictions on the range of definition;

The periodicity of all trigonometric functions, because We have already noted the presence of the smallest non-zero argument, the addition of which does not change the value of the function. This argument is called the period of the function and is denoted by the letter . For sine/cosine and tangent/cotangent these periods are different.

Consider the function:

1) Scope of definition;

2) Value range ;

3) The function is odd ;

Let's build a graph of the function. In this case, it is convenient to begin the construction with an image of the area that limits the graph from above by the number 1 and from below by the number , which is associated with the range of values ​​of the function. In addition, for construction it is useful to remember the values ​​of the sines of several main table angles, for example, that this will allow you to build the first full “wave” of the graph and then redraw it to the right and left, taking advantage of the fact that the picture will be repeated with an offset by a period, i.e. on .

Now let's look at the function:

The main properties of this function:

1) Scope of definition;

2) Value range ;

3) Even function This implies that the graph of the function is symmetrical about the ordinate;

4) The function is not monotonic throughout its entire domain of definition;

Let's build a graph of the function. As when constructing a sine, it is convenient to start with an image of the area that limits the graph at the top with the number 1 and at the bottom with the number , which is associated with the range of values ​​of the function. We will also plot the coordinates of several points on the graph, for which we need to remember the values ​​of the cosines of several main table angles, for example, that with the help of these points we can build the first full “wave” of the graph and then redraw it to the right and left, taking advantage of the fact that the picture will repeat with a period shift, i.e. on .

Let's move on to the function:

The main properties of this function:

1) Domain except , where . We have already indicated in previous lessons that it does not exist. This statement can be generalized by considering the tangent period;

2) Range of values, i.e. tangent values ​​are not limited;

3) The function is odd ;

4) The function increases monotonically within its so-called tangent branches, which we will now see in the figure;

5) The function is periodic with a period

Let's build a graph of the function. In this case, it is convenient to begin the construction by depicting the vertical asymptotes of the graph at points that are not included in the definition domain, i.e. etc. Next, we depict the tangent branches inside each of the strips formed by the asymptotes, pressing them to the left asymptote and to the right one. At the same time, do not forget that each branch increases monotonically. We depict all branches the same way, because the function has a period equal to . This can be seen from the fact that each branch is obtained by shifting the neighboring one along the abscissa axis.

And we finish with a look at the function:

The main properties of this function:

1) Domain except , where . From the table of values ​​of trigonometric functions, we already know that it does not exist. This statement can be generalized by considering the cotangent period;

2) Range of values, i.e. cotangent values ​​are not limited;

3) The function is odd ;

4) The function decreases monotonically within its branches, which are similar to the tangent branches;

5) The function is periodic with a period

Let's build a graph of the function. In this case, as for the tangent, it is convenient to begin the construction by depicting the vertical asymptotes of the graph at points that are not included in the definition area, i.e. etc. Next, we depict the branches of the cotangent inside each of the stripes formed by the asymptotes, pressing them to the left asymptote and to the right one. In this case, we take into account that each branch decreases monotonically. We depict all branches similarly to the tangent in the same way, because the function has a period equal to .

Separately, it should be noted that trigonometric functions with complex arguments may have a non-standard period. We are talking about functions of the form:

Their period is equal. And about the functions:

Their period is equal.

As you can see, to calculate a new period, the standard period is simply divided by the factor in the argument. It does not depend on other modifications of the function.

You can understand in more detail and understand where these formulas come from in the lesson about constructing and transforming graphs of functions.

We have come to one of the most important parts of the topic “Trigonometry”, which we will devote to solving trigonometric equations. The ability to solve such equations is important, for example, when describing oscillatory processes in physics. Let's imagine that you have driven a few laps in a go-kart in a sports car; solving a trigonometric equation will help you determine how long you have been racing depending on the position of the car on the track.

Let's write the simplest trigonometric equation:

The solution to such an equation is the arguments whose sine is equal to . But we already know that due to the periodicity of the sine, there is an infinite number of such arguments. Thus, the solution to this equation will be, etc. The same applies to solving any other simple trigonometric equation; there will be an infinite number of them.

Trigonometric equations are divided into several main types. Separately, we should dwell on the simplest ones, because everything else comes down to them. There are four such equations (according to the number of basic trigonometric functions). General solutions are known for them; they must be remembered.

The simplest trigonometric equations and their general solutions look like this:

Please note that the values ​​of sine and cosine must take into account the limitations known to us. If, for example, then the equation has no solutions and the specified formula should not be applied.

In addition, the specified root formulas contain a parameter in the form of an arbitrary integer. In the school curriculum, this is the only case when the solution to an equation without a parameter contains a parameter. This arbitrary integer shows that it is possible to write down an infinite number of roots of any of the above equations simply by substituting all the integers in turn.

You can get acquainted with the detailed derivation of these formulas by repeating the chapter “Trigonometric Equations” in the 10th grade algebra program.

Separately, it is necessary to pay attention to solving special cases of the simplest equations with sine and cosine. These equations look like:

Formulas for finding general solutions should not be applied to them. Such equations are most conveniently solved using the trigonometric circle, which gives a simpler result than general solution formulas.

For example, the solution to the equation is . Try to get this answer yourself and solve the remaining equations indicated.

In addition to the most common type of trigonometric equations indicated, there are several more standard ones. We list them taking into account those that we have already indicated:

1) Protozoa, For example, ;

2) Special cases of the simplest equations, For example, ;

3) Equations with complex argument, For example, ;

4) Equations reduced to their simplest by taking out a common factor, For example, ;

5) Equations reduced to their simplest by transforming trigonometric functions, For example, ;

6) Equations reduced to their simplest by substitution, For example, ;

7) Homogeneous equations, For example, ;

8) Equations that can be solved using the properties of functions, For example, . Don’t be alarmed by the fact that there are two variables in this equation; it solves itself;

As well as equations that are solved using various methods.

In addition to solving trigonometric equations, you must be able to solve their systems.

The most common types of systems are:

1) In which one of the equations is power, For example, ;

2) Systems of simple trigonometric equations, For example, .

In today's lesson we looked at the basic trigonometric functions, their properties and graphs. We also got acquainted with the general formulas for solving the simplest trigonometric equations, indicated the main types of such equations and their systems.

In the practical part of the lesson, we will examine methods for solving trigonometric equations and their systems.

Box 1.Solving special cases of the simplest trigonometric equations.

As we already said in the main part of the lesson, special cases of trigonometric equations with sine and cosine of the form:

have simpler solutions than those given by the general solution formulas.

A trigonometric circle is used for this. Let us analyze the method for solving them using the example of the equation.

Let us depict on the trigonometric circle the point at which the cosine value is zero, which is also the coordinate along the abscissa axis. As you can see, there are two such points. Our task is to indicate what the angle that corresponds to these points on the circle is equal to.

We start counting from the positive direction of the abscissa axis (cosine axis) and when setting the angle we get to the first depicted point, i.e. one solution would be this angle value. But we are still satisfied with the angle that corresponds to the second point. How to get into it?

Task No. 1

The logic is simple: we will do as we did before, regardless of the fact that now trigonometric functions have a more complex argument!

If we were to solve an equation of the form:

Then we would write down the following answer:

Or (since)

But now our role is played by this expression:

Then we can write:

Our goal with you is to make sure that the left side stands simply, without any “impurities”!

Let's gradually get rid of them!

First, let’s remove the denominator at: to do this, multiply our equality by:

Now let's get rid of it by dividing both parts:

Now let's get rid of the eight:

The resulting expression can be written as 2 series of solutions (by analogy with a quadratic equation, where we either add or subtract the discriminant)

We need to find the largest negative root! It is clear that we need to sort through.

Let's look at the first episode first:

It is clear that if we take, then as a result we will receive positive numbers, but they do not interest us.

So you need to take it negative. Let be.

When the root will be narrower:

And we need to find the greatest negative!! This means that going in the negative direction no longer makes sense here. And the largest negative root for this series will be equal to.

Now let's look at the second series:

And again we substitute: , then:

Not interested!

Then it makes no sense to increase any more! Let's reduce it! Let then:

Fits!

Let be. Then

Then - the greatest negative root!

Answer:

Task No. 2

We solve again, regardless of the complex cosine argument:

Now we express again on the left:

Multiply both sides by

Divide both sides by

All that remains is to move it to the right, changing its sign from minus to plus.

We again get 2 series of roots, one with and the other with.

We need to find the largest negative root. Let's look at the first episode:

It is clear that we will get the first negative root at, it will be equal to and will be the largest negative root in 1 series.

For the second series

The first negative root will also be obtained at and will be equal to. Since, then is the largest negative root of the equation.

Answer: .

Task No. 3

We solve, regardless of the complex tangent argument.

Now, it doesn’t seem complicated, right?

As before, we express on the left side:

Well, that’s great, there’s only one series of roots here! Let's find the largest negative again.

It is clear that it turns out if you put it down. And this root is equal.

Answer:

Now try to solve the following problems yourself.

Homework or 3 tasks to solve independently.

1. Resolve the equation.
   
2. Resolve the equation.
   In the answer to the pi-shi-th-the-smallest-possible root.
3. Resolve the equation.
   In the answer to the pi-shi-th-the-smallest-possible root.

Ready? Let's check. I will not describe in detail the entire solution algorithm; it seems to me that it has already received enough attention above.

Well, is everything right? Oh, those nasty sinuses, there’s always some kind of trouble with them!

Well, now you can solve simple trigonometric equations!

Check out the solutions and answers:

Task No. 1

Let's express

The smallest positive root is obtained if we put, since, then

Answer:

Task No. 2

The smallest positive root is obtained at.

It will be equal.

Answer: .

Task No. 3

When we get, when we have.

Answer: .

This knowledge will help you solve many problems that you will encounter in the exam.

If you are applying for a “5” rating, then you just need to proceed to reading the article for mid-level which will be devoted to solving more complex trigonometric equations (task C1).

AVERAGE LEVEL

In this article I will describe solving more complex trigonometric equations and how to select their roots. Here I will draw on the following topics:

1. Trigonometric equations for beginner level (see above).

More complex trigonometric equations are the basis for advanced problems. They require both solving the equation itself in general form and finding the roots of this equation belonging to a certain given interval.

Solving trigonometric equations comes down to two subtasks:

1. Solving the equation
2. Root selection

It should be noted that the second is not always required, but in most examples selection is still required. But if it is not required, then we can sympathize with you - this means that the equation is quite complex in itself.

My experience in analyzing C1 problems shows that they are usually divided into the following categories.

Four categories of tasks of increased complexity (formerly C1)

1. Equations that reduce to factorization.
2. Equations reduced to form.
3. Equations solved by changing a variable.
4. Equations that require additional selection of roots due to irrationality or denominator.

To put it simply: if you get caught one of the equations of the first three types, then consider yourself lucky. For them, as a rule, you additionally need to select roots belonging to a certain interval.

If you come across a type 4 equation, then you are less lucky: you need to tinker with it longer and more carefully, but quite often it does not require additional selection of roots. Nevertheless, I will analyze this type of equations in the next article, and this one I will devote to solving equations of the first three types.

Equations that reduce to factorization

The most important thing you need to remember to solve this type of equation is

As practice shows, as a rule, this knowledge is sufficient. Let's look at some examples:

Example 1. Equation reduced to factorization using the reduction and double angle sine formulas

• Resolve the equation
• Find all the roots of this equation that lie above the cut

Here, as I promised, the reduction formulas work:

Then my equation will look like this:

Then my equation will take the following form:

A short-sighted student might say: now I’ll reduce both sides by, get the simplest equation and enjoy life! And he will be bitterly mistaken!

REMEMBER: YOU CAN NEVER REDUCE BOTH SIDES OF A TRIGONOMETRIC EQUATION BY A FUNCTION CONTAINING AN UNKNOWN! SO YOU LOSE YOUR ROOTS!

So what to do? Yes, it’s simple, move everything to one side and take out the common factor:

Well, we factored it into factors, hurray! Now let's decide:

The first equation has roots:

And the second:

This completes the first part of the problem. Now you need to select the roots:

The gap is like this:

Or it can also be written like this:

Well, let's take the roots:

First, let's work with the first episode (and it's simpler, to say the least!)

Since our interval is entirely negative, there is no need to take non-negative ones, they will still give non-negative roots.

Let's take it, then - it's too much, it doesn't hit.

Let it be, then - I didn’t hit it again.

One more try - then - yes, I got it! The first root has been found!

I shoot again: then I hit again!

Well, one more time: : - this is already a flight.

So from the first series there are 2 roots belonging to the interval: .

We are working with the second series (we are building to the power according to the rule):

Undershoot!

Missing it again!

Missing it again!

Got it!

Flight!

Thus, my interval has the following roots:

This is the algorithm we will use to solve all other examples. Let's practice together with one more example.

Example 2. Equation reduced to factorization using reduction formulas

• Solve the equation

Solution:

Again the notorious reduction formulas:

Don't try to cut back again!

The first equation has roots:

And the second:

Now again the search for roots.

I’ll start with the second episode, I already know everything about it from the previous example! Look and make sure that the roots belonging to the interval are as follows:

Now the first episode and it’s simpler:

If - suitable

If that's fine too

If it’s already a flight.

Then the roots will be as follows:

Independent work. 3 equations.

Well, is the technique clear to you? Does solving trigonometric equations not seem so difficult anymore? Then quickly solve the following problems yourself, and then we will solve other examples:

1. Solve the equation
   Find all the roots of this equation that lie above the interval.
2. Resolve the equation
   Indicate the roots of the equation that lie above the cut
3. Resolve the equation
   Find all the roots of this equation that lie between them.

Equation 1.

And again the reduction formula:

First series of roots:

Second series of roots:

We begin selection for the gap

Answer: , .

Equation 2. Checking independent work.

Quite a tricky grouping into factors (I’ll use the double angle sine formula):

then or

This is a general solution. Now we need to select the roots. The trouble is that we cannot tell the exact value of an angle whose cosine is equal to one quarter. Therefore, I can’t just get rid of the arc cosine - such a shame!

What I can do is figure out that so, so, then.

Let's create a table: interval:

Well, through painful searches we came to the disappointing conclusion that our equation has one root on the indicated interval: \displaystyle arccos\frac(1)(4)-5\pi

Equation 3: Independent work test.

A frightening looking equation. However, it can be solved quite simply by applying the double angle sine formula:

Let's reduce it by 2:

Let's group the first term with the second and the third with the fourth and take out the common factors:

It is clear that the first equation has no roots, and now let’s consider the second:

In general, I was going to dwell a little later on solving such equations, but since it turned up, there’s nothing to do, I have to solve it...

Equations of the form:

This equation is solved by dividing both sides by:

Thus, our equation has a single series of roots:

We need to find those that belong to the interval: .

Let's build a table again, as I did earlier:

Answer: .

Equations reduced to the form:

Well, now it’s time to move on to the second portion of equations, especially since I’ve already spilled the beans on what the solution to trigonometric equations of a new type consists of. But it is worth repeating that the equation is of the form

Solved by dividing both sides by cosine:

1. Resolve the equation
   Indicate the roots of the equation that lie above the cut.
2. Resolve the equation
   Indicate the roots of the equation that lie between them.

Example 1.

The first one is quite simple. Move to the right and apply the double angle cosine formula:

Yeah! Equation of the form: . I divide both parts by

We do root screening:

Gap:

Answer:

Example 2.

Everything is also quite trivial: let’s open the brackets on the right:

Basic trigonometric identity:

Sine of double angle:

Finally we get:

Root screening: interval.

Answer: .

Well, how do you like the technique, isn’t it too complicated? I hope not. We can immediately make a reservation: in their pure form, equations that immediately reduce to an equation for the tangent are quite rare. Typically, this transition (division by cosine) is only part of a more complex problem. Here's an example for you to practice:

• Resolve the equation
• Find all the roots of this equation that lie above the cut.

Let's check:

The equation can be solved immediately; it is enough to divide both sides by:

Root screening:

Answer: .

One way or another, we have yet to encounter equations of the type that we have just examined. However, it’s too early for us to call it a day: there’s still one more “layer” of equations left that we haven’t sorted out. So:

Solving trigonometric equations by changing variables

Everything is transparent here: we look closely at the equation, simplify it as much as possible, make a substitution, solve it, make a reverse substitution! In words everything is very easy. Let's see in action:

Example.

• Solve the equation: .
• Find all the roots of this equation that lie above the cut.

Well, here the replacement itself suggests itself to us!

Then our equation will turn into this:

The first equation has roots:

And the second one is like this:

Now let's find the roots belonging to the interval

Answer: .

Let's look at a slightly more complex example together:

• Resolve the equation
• Indicate the roots of the given equation, lying above-lying between them.

Here the replacement is not immediately visible, moreover, it is not very obvious. Let's first think: what can we do?

We can, for example, imagine

And at the same time

Then my equation will take the form:

And now attention, focus:

Let's divide both sides of the equation by:

Suddenly you and I have a quadratic equation relative! Let's make a replacement, then we get:

The equation has the following roots:

Unpleasant second series of roots, but nothing can be done! We select roots in the interval.

We also need to consider that

Since and, then

Answer:

To reinforce this before you solve the problems yourself, here’s another exercise for you:

• Resolve the equation
• Find all the roots of this equation that lie between them.

Here you need to keep your eyes open: we now have denominators that can be zero! Therefore, you need to be especially attentive to the roots!

First of all, I need to rearrange the equation so that I can make a suitable substitution. I can’t think of anything better now than to rewrite the tangent in terms of sine and cosine:

Now I will move from cosine to sine using the basic trigonometric identity:

And finally, I’ll bring everything to a common denominator:

Now I can move on to the equation:

But at (that is, at).

Now everything is ready for replacement:

Then or

However, note that if, then at the same time!

Who suffers from this? The problem with the tangent is that it is not defined when the cosine is equal to zero (division by zero occurs).

Thus, the roots of the equation are:

Now we sift out the roots in the interval:

	- fits
	- overkill

Thus, our equation has a single root on the interval, and it is equal.

You see: the appearance of a denominator (just like the tangent, leads to certain difficulties with the roots! Here you need to be more careful!).

Well, you and I have almost finished analyzing trigonometric equations; there is very little left - to solve two problems on your own. Here they are.

1. Solve the equation
   Find all the roots of this equation that lie above the cut.
2. Resolve the equation
   Indicate the roots of this equation, located above the cut.

Decided? Isn't it very difficult? Let's check:

1. We work according to the reduction formulas:

   Substitute into the equation:

   Let's rewrite everything through cosines to make it easier to make the replacement:

   Now it's easy to make a replacement:

   It is clear that is an extraneous root, since the equation has no solutions. Then:

   We are looking for the roots we need in the interval

   Answer: .

2. 
   Here the replacement is immediately visible:

   Then or

   	- fits!	- fits!
   	- fits!	- fits!
   	- a lot of!	- also a lot!

   Answer:

Well, that's it now! But solving trigonometric equations does not end there; we are left behind in the most difficult cases: when the equations contain irrationality or various kinds of “complex denominators.” We will look at how to solve such tasks in an article for an advanced level.

ADVANCED LEVEL

In addition to the trigonometric equations discussed in the previous two articles, we will consider another class of equations that require even more careful analysis. These trigonometric examples contain either irrationality or a denominator, which makes their analysis more difficult. However, you may well encounter these equations in Part C of the exam paper. However, every cloud has a silver lining: for such equations, as a rule, the question of which of its roots belongs to a given interval is no longer raised. Let's not beat around the bush, but let's go straight to trigonometric examples.

Example 1.

Solve the equation and find the roots that belong to the segment.

Solution:

We have a denominator that should not be equal to zero! Then solving this equation is the same as solving the system

Let's solve each of the equations:

And now the second one:

Now let's look at the series:

It is clear that this option does not suit us, since in this case our denominator is reset to zero (see the formula for the roots of the second equation)

If, then everything is in order, and the denominator is not zero! Then the roots of the equation are as follows: , .

Now we select the roots belonging to the interval.

	- not suitable	- fits
	- fits	- fits
	overkill	overkill

Then the roots are as follows:

You see, even the appearance of a small disturbance in the form of the denominator significantly affected the solution of the equation: we discarded a series of roots that nullified the denominator. Things can get even more complicated if you come across trigonometric examples that are irrational.

Example 2.

Solve the equation:

Solution:

Well, at least you don’t have to take away the roots, and that’s good! Let's first solve the equation, regardless of irrationality:

So, is that all? No, alas, it would be too easy! We must remember that only non-negative numbers can appear under the root. Then:

The solution to this inequality is:

Now it remains to find out whether part of the roots of the first equation inadvertently ended up where the inequality does not hold.

To do this, you can again use the table:

	: , But	No!
		Yes!
		Yes!

Thus, one of my roots “fell out”! It turns out if you put it down. Then the answer can be written as follows:

Answer:

You see, the root requires even more attention! Let's make it more complicated: let now I have a trigonometric function under my root.

Example 3.

As before: first we will solve each one separately, and then we will think about what we have done.

Now the second equation:

Now the most difficult thing is to find out whether negative values ​​​​are obtained under the arithmetic root if we substitute the roots from the first equation there:

The number must be understood as radians. Since a radian is approximately degrees, then radians are on the order of degrees. This is the corner of the second quarter. What is the sign of the cosine of the second quarter? Minus. What about sine? Plus. So what can we say about the expression:

It's less than zero!

This means that it is not the root of the equation.

Now it's time.

Let's compare this number with zero.

Cotangent is a function decreasing in 1 quarter (the smaller the argument, the greater the cotangent). radians are approximately degrees. In the same time

since, then, and therefore
,

Answer: .

Could it get any more complicated? Please! It will be more difficult if the root is still a trigonometric function, and the second part of the equation is again a trigonometric function.

The more trigonometric examples the better, see below:

Example 4.

The root is not suitable due to the limited cosine

Now the second one:

At the same time, by definition of a root:

We need to remember the unit circle: namely, those quarters where the sine is less than zero. What are these quarters? Third and fourth. Then we will be interested in those solutions of the first equation that lie in the third or fourth quarter.

The first series gives roots lying at the intersection of the third and fourth quarters. The second series - diametrically opposite to it - gives rise to roots lying on the border of the first and second quarters. Therefore, this series is not suitable for us.

Answer: ,

And again trigonometric examples with "difficult irrationality". Not only do we have the trigonometric function under the root again, but now it’s also in the denominator!

Example 5.

Well, nothing can be done - we do as before.

Now we work with the denominator:

I don’t want to solve the trigonometric inequality, so I’ll do something cunning: I’ll take and substitute my series of roots into the inequality:

If - is even, then we have:

since all angles of the view lie in the fourth quarter. And again the sacred question: what is the sign of the sine in the fourth quarter? Negative. Then the inequality

If -odd, then:

In which quarter does the angle lie? This is the corner of the second quarter. Then all the corners are again the corners of the second quarter. The sine there is positive. Just what you need! So the series:

Fits!

We deal with the second series of roots in the same way:

We substitute into our inequality:

If - even, then

First quarter corners. The sine there is positive, which means the series is suitable. Now if - odd, then:

fits too!

Well, now we write down the answer!

Answer:

Well, this was perhaps the most labor-intensive case. Now I offer you problems to solve on your own.

Training

1. Solve and find all the roots of the equation that belong to the segment.

Solutions:

1. 
   First equation:
   or
   ODZ of the root:

   Second equation:

   Selection of roots that belong to the interval

   Answer:

Or
or
But

Let's consider: . If - even, then
- doesn't fit!
If - odd, : - suitable!
This means that our equation has the following series of roots:
or
Selection of roots in the interval:

	- not suitable	- fits
	- fits	- a lot of
	- fits	a lot of

Answer: , .

Or
Since, then the tangent is not defined. We immediately discard this series of roots!

Second part:

At the same time, according to DZ it is required that

We check the roots found in the first equation:

If the sign:

First quarter angles where the tangent is positive. Doesn't fit!
If the sign:

Fourth quarter corner. There the tangent is negative. Fits. We write down the answer:

Answer: , .

We have looked at complex trigonometric examples together in this article, but you should solve the equations yourself.

SUMMARY AND BASIC FORMULAS

A trigonometric equation is an equation in which the unknown is strictly under the sign of the trigonometric function.

There are two ways to solve trigonometric equations:

The first way is using formulas.

The second way is through the trigonometric circle.

Allows you to measure angles, find their sines, cosines, etc.