I will solve the exam trigonometric equations. Solving trigonometric equations and methods for selecting roots on a given interval. A selection of assignments from previous years

27. Let's select roots on a given segment

28. 10. Solve the equation 2sin2x =4cos x –sinx+1 Find its roots on the interval [/2; 3/2]

29. Let's select roots using a circle

A) Solve equation 2(\sin x-\cos x)=tgx-1.

b) \left[ \frac(3\pi )2;\,3\pi \right].

Solution

A) Opening the brackets and moving all the terms to the left side, we get the equation 1+2 \sin x-2 \cos x-tg x=0. Considering that \cos x \neq 0, the term 2 \sin x can be replaced by 2 tan x \cos x, we obtain the equation 1+2 tg x \cos x-2 \cos x-tg x=0, which by grouping can be reduced to the form (1-tg x)(1-2 \cos x)=0.

1) 1-tg x=0, tan x=1, x=\frac\pi 4+\pi n, n \in \mathbb Z;

2) 1-2 \cos x=0, \cos x=\frac12, x=\pm \frac\pi 3+2\pi n, n \in \mathbb Z.

b) Using the number circle, select the roots belonging to the interval \left[ \frac(3\pi )2;\, 3\pi \right].

x_1=\frac\pi 4+2\pi =\frac(9\pi )4,

x_2=\frac\pi 3+2\pi =\frac(7\pi )3,

x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.

Answer

A) \frac\pi 4+\pi n, \pm\frac\pi 3+2\pi n, n \in \mathbb Z;

b) \frac(5\pi )3, \frac(7\pi )3, \frac(9\pi )4.

Condition

A) Solve the equation (2\sin ^24x-3\cos 4x)\cdot \sqrt (tgx)=0.

b) Indicate the roots of this equation that belong to the interval \left(0;\,\frac(3\pi )2\right] ;

Solution

A) ODZ: \begin(cases) tgx\geqslant 0\\x\neq \frac\pi 2+\pi k,k \in \mathbb Z. \end(cases)

The original equation on the ODZ is equivalent to a set of equations

\left[\!\!\begin(array)(l) 2 \sin ^2 4x-3 \cos 4x=0,\\tg x=0. \end(array)\right.

Let's solve the first equation. To do this we will make a replacement \cos 4x=t, t \in [-1; 1]. Then \sin^24x=1-t^2. We get:

2(1-t^2)-3t=0,

2t^2+3t-2=0,

t_1=\frac12, t_2=-2, t_2\notin [-1; 1].

\cos 4x=\frac12,

4x=\pm\frac\pi 3+2\pi n,

x=\pm \frac\pi (12)+\frac(\pi n)2, n \in \mathbb Z.

Let's solve the second equation.

tg x=0,\, x=\pi k, k \in \mathbb Z.

Using the unit circle, we find solutions that satisfy the ODZ.

The “+” sign marks the 1st and 3rd quarters, in which tg x>0.

We get: x=\pi k, k \in \mathbb Z; x=\frac\pi (12)+\pi n, n \in \mathbb Z; x=\frac(5\pi )(12)+\pi m, m \in \mathbb Z.

b) Let's find the roots belonging to the interval \left(0;\,\frac(3\pi )2\right].

x=\frac\pi (12), x=\frac(5\pi )(12); x=\pi ; x=\frac(13\pi )(12); x=\frac(17\pi )(12).

Answer

A) \pi k, k \in \mathbb Z; \frac\pi (12)+\pi n, n \in \mathbb Z; \frac(5\pi )(12)+\pi m, m \in \mathbb Z.

b) \pi; \frac\pi (12); \frac(5\pi )(12); \frac(13\pi )(12); \frac(17\pi )(12).

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

A) Solve the equation: \cos ^2x+\cos ^2\frac\pi 6=\cos ^22x+\sin ^2\frac\pi 3;

b) List all roots belonging to the interval \left(\frac(7\pi )2;\,\frac(9\pi )2\right].

Solution

A) Because \sin \frac\pi 3=\cos \frac\pi 6, That \sin ^2\frac\pi 3=\cos ^2\frac\pi 6, This means that the given equation is equivalent to the equation \cos^2x=\cos ^22x, which, in turn, is equivalent to the equation \cos^2x-\cos ^2 2x=0.

But \cos ^2x-\cos ^22x= (\cos x-\cos 2x)\cdot (\cos x+\cos 2x) And

\cos 2x=2 \cos ^2 x-1, so the equation becomes

(\cos x-(2 \cos ^2 x-1))\,\cdot(\cos x+(2 \cos ^2 x-1))=0,

(2 \cos ^2 x-\cos x-1)\,\cdot (2 \cos ^2 x+\cos x-1)=0.

Then either 2 \cos ^2 x-\cos x-1=0, or 2 \cos ^2 x+\cos x-1=0.

Solving the first equation as a quadratic equation for \cos x, we get:

(\cos x)_(1,2)=\frac(1\pm\sqrt 9)4=\frac(1\pm3)4. Therefore either \cos x=1 or \cos x=-\frac12. If \cos x=1, then x=2k\pi , k \in \mathbb Z. If \cos x=-\frac12, That x=\pm \frac(2\pi )3+2s\pi , s \in \mathbb Z.

Similarly, solving the second equation, we get either \cos x=-1 or \cos x=\frac12. If \cos x=-1, then the roots x=\pi +2m\pi , m \in \mathbb Z. If \cos x=\frac12, That x=\pm \frac\pi 3+2n\pi , n \in \mathbb Z.

Let's combine the solutions obtained:

x=m\pi , m \in \mathbb Z; x=\pm \frac\pi 3 +s\pi , s \in \mathbb Z.

b) Let's select the roots that fall within a given interval using a number circle.

We get: x_1 =\frac(11\pi )3, x_2=4\pi , x_3 =\frac(13\pi )3.

Answer

A) m\pi, m\in \mathbb Z; \pm \frac\pi 3 +s\pi , s \in \mathbb Z;

b) \frac(11\pi )3, 4\pi , \frac(13\pi )3.

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

A) Solve the equation 10\cos ^2\frac x2=\frac(11+5ctg\left(\dfrac(3\pi )2-x\right) )(1+tgx).

b) Indicate the roots of this equation that belong to the interval \left(-2\pi ; -\frac(3\pi )2\right).

Solution

A) 1. According to the reduction formula, ctg\left(\frac(3\pi )2-x\right) =tgx. The domain of definition of the equation will be such values ​​of x such that \cos x \neq 0 and tan x \neq -1. Let's transform the equation using the double angle cosine formula 2 \cos ^2 \frac x2=1+\cos x. We get the equation: 5(1+\cos x) =\frac(11+5tgx)(1+tgx).

notice, that \frac(11+5tgx)(1+tgx)= \frac(5(1+tgx)+6)(1+tgx)= 5+\frac(6)(1+tgx), so the equation becomes: 5+5 \cos x=5 +\frac(6)(1+tgx). From here \cos x =\frac(\dfrac65)(1+tgx), \cos x+\sin x =\frac65.

2. Transform \sin x+\cos x using the reduction formula and the sum of cosines formula: \sin x=\cos \left(\frac\pi 2-x\right), \cos x+\sin x= \cos x+\cos \left(\frac\pi 2-x\right)= 2\cos \frac\pi 4\cos \left(x-\frac\pi 4\right)= \sqrt 2\cos \left(x-\frac\pi 4\right) = \frac65.

From here \cos \left(x-\frac\pi 4\right) =\frac(3\sqrt 2)5. Means, x-\frac\pi 4= arc\cos \frac(3\sqrt 2)5+2\pi k, k \in \mathbb Z,

or x-\frac\pi 4= -arc\cos \frac(3\sqrt 2)5+2\pi t, t \in \mathbb Z.

That's why x=\frac\pi 4+arc\cos \frac(3\sqrt 2)5+2\pi k,k \in \mathbb Z,

or x =\frac\pi 4-arc\cos \frac(3\sqrt 2)5+2\pi t,t \in \mathbb Z.

The found values ​​of x belong to the domain of definition.

b) Let us first find out where the roots of the equation fall at k=0 and t=0. These will be numbers accordingly a=\frac\pi 4+arccos \frac(3\sqrt 2)5 And b=\frac\pi 4-arccos \frac(3\sqrt 2)5.

1. Let us prove the auxiliary inequality:

\frac(\sqrt 2)(2)<\frac{3\sqrt 2}2<1.

Really, \frac(\sqrt 2)(2)=\frac(5\sqrt 2)(10)<\frac{6\sqrt2}{10}=\frac{3\sqrt2}{5}.

Note also that \left(\frac(3\sqrt 2)5\right) ^2=\frac(18)(25)<1^2=1, Means \frac(3\sqrt 2)5<1.

2. From inequalities (1) By the arc cosine property we get:

arccos 1

0

From here \frac\pi 4+0<\frac\pi 4+arc\cos \frac{3\sqrt 2}5<\frac\pi 4+\frac\pi 4,

0<\frac\pi 4+arccos \frac{3\sqrt 2}5<\frac\pi 2,

0

Likewise, -\frac\pi 4

0=\frac\pi 4-\frac\pi 4<\frac\pi 4-arccos \frac{3\sqrt 2}5< \frac\pi 4<\frac\pi 2,

0

For k=-1 and t=-1 we obtain the roots of the equation a-2\pi and b-2\pi.

\Bigg(a-2\pi =-\frac74\pi +arccos \frac(3\sqrt 2)5,\, b-2\pi =-\frac74\pi -arccos \frac(3\sqrt 2)5\Bigg). Wherein -2\pi

2\pi This means that these roots belong to the given interval \left(-2\pi , -\frac(3\pi )2\right).

For other values ​​of k and t, the roots of the equation do not belong to the given interval.

Indeed, if k\geqslant 1 and t\geqslant 1, then the roots are greater than 2\pi. If k\leqslant -2 and t\leqslant -2, then the roots are smaller -\frac(7\pi )2.

Answer

A) \frac\pi4\pm arccos\frac(3\sqrt2)5+2\pi k, k\in\mathbb Z;

b) -\frac(7\pi)4\pm arccos\frac(3\sqrt2)5.

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

A) Solve the equation \sin \left(\frac\pi 2+x\right) =\sin (-2x).

b) Find all the roots of this equation that belong to the interval ;

Solution

A) Let's transform the equation:

\cos x =-\sin 2x,

\cos x+2 \sin x \cos x=0,

\cos x(1+2 \sin x)=0,

\cos x=0,

x =\frac\pi 2+\pi n, n\in \mathbb Z;

1+2 \sin x=0,

\sin x=-\frac12,

x=(-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z.

b) We find the roots belonging to the segment using the unit circle.

The indicated interval contains a single number \frac\pi 2.

Answer

A) \frac\pi 2+\pi n, n \in \mathbb Z; (-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z;

b) \frac\pi 2.

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

Means, \sin x \neq 1.

Divide both sides of the equation by a factor (\sin x-1), different from zero. We get the equation \frac 1(1+\cos 2x)=\frac 1(1+\cos (\pi +x)), or equation 1+\cos 2x=1+\cos (\pi +x). Applying the reduction formula on the left side and the reduction formula on the right, we obtain the equation 2 \cos ^2 x=1-\cos x. This equation is by substitution \cos x=t, Where -1 \leqslant t \leqslant 1 reduce it to square: 2t^2+t-1=0, whose roots t_1=-1 And t_2=\frac12. Returning to the variable x, we get \cos x = \frac12 or \cos x=-1, where x=\frac \pi 3+2\pi m, m \in \mathbb Z, x=-\frac \pi 3+2\pi n, n \in \mathbb Z, x=\pi +2\pi k, k \in \mathbb Z.

b) Let's solve inequalities

1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 ,

2) -\frac(3\pi )2 \leqslant -\frac \pi 3+2\pi n \leqslant -\frac \pi (2,)

3) -\frac(3\pi )2 \leqslant \pi+2\pi k \leqslant -\frac \pi 2 , m, n, k \in \mathbb Z.

1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 , -\frac32\leqslant \frac13+2m \leqslant -\frac12 -\frac(11)6 \leqslant 2m\leqslant -\frac56 , -\frac(11)(12) \leqslant m \leqslant -\frac5(12).

\left [-\frac(11)(12);-\frac5(12)\right].

2) -\frac (3\pi) 2 \leqslant -\frac(\pi )3+2\pi n \leqslant -\frac(\pi )(2), -\frac32 \leqslant -\frac13 +2n \leqslant -\frac12 , -\frac76 \leqslant 2n \leqslant -\frac1(6), -\frac7(12) \leqslant n \leqslant -\frac1(12).

There are no integers in the range \left[ -\frac7(12) ; -\frac1(12)\right].

3) -\frac(3\pi )2 \leqslant \pi +2\pi k\leqslant -\frac(\pi )2, -\frac32 \leqslant 1+2k\leqslant -\frac12, -\frac52 \leqslant 2k \leqslant -\frac32, -\frac54 \leqslant k \leqslant -\frac34.

This inequality is satisfied by k=-1, then x=-\pi.

Answer

A) \frac \pi 3+2\pi m; -\frac \pi 3+2\pi n; \pi +2\pi k, m, n, k \in \mathbb Z;

b) -\pi .

a) Solve the equation: .

b) Find all the roots of this equation that belong to the segment.

The solution of the problem

This lesson discusses an example of solving a trigonometric equation, which can be used as an example for solving problems of type C1 when preparing for the Unified State Exam in mathematics.

First of all, the scope of the function is determined - all valid values ​​of the argument. Then, during the solution, the trigonometric sine function is converted to cosine using the reduction formula. Next, all terms of the equation are transferred to its left side, where the common factor is taken out of brackets. Each factor is equal to zero, which allows us to determine the roots of the equation. Then, using the method of turns, the roots belonging to a given segment are determined. To do this, on the constructed unit circle, a turn is marked from the left border of a given segment to the right. Next, the found roots on the unit circle are connected by segments to its center and the points at which these segments intersect the turn are determined. These intersection points are the desired answer to the second part of the problem.

The purpose of the lesson:

A) strengthen the ability to solve simple trigonometric equations;

b) teach how to select roots of trigonometric equations from a given interval

During the classes.

a) Checking homework: the class is given advanced homework - solve an equation and find a way to select roots from a given interval.

1)cos x= -0.5, where xI [- ]. Answer:.

2) sin x= , where xI . Answer: ; .

3)cos 2 x= -, where xI. Answer:

Students write down the solution on the board, some using a graph, others using the selection method.

At this time class works orally.

Find the meaning of the expression:

a) tg – sin + cos + sin. Answer: 1.

b) 2arccos 0 + 3 arccos 1. Answer: ?

c) arcsin + arcsin. Answer:.

d) 5 arctg (-) – arccos (-). Answer:-.

– Let’s check your homework, open your notebooks with homework.

Some of you found the solution using the selection method, and some using the graph.

2. Conclusion about ways to solve these tasks and statement of the problem, i.e., communication of the topic and purpose of the lesson.

– a) It is difficult to solve using selection if a large interval is given.

– b) The graphical method does not give accurate results, requires verification, and takes a lot of time.

– Therefore, there must be at least one more method, the most universal one - let’s try to find it. So, what are we going to do in class today? (Learn to choose the roots of a trigonometric equation on a given interval.)

– Example 1. (Student goes to the board)

cos x= -0.5, where xI [- ].

Question: What determines the answer to this task? (From the general solution of the equation. Let us write the solution in general form). The solution is written on the board

x = + 2?k, where k R.

– Let’s write this solution in the form of a set:

– In your opinion, in what notation of the solution is it convenient to choose roots on the interval? (from the second entry). But this is again a selection method. What do we need to know to get the right answer? (You need to know the values ​​of k).

(Let's create a mathematical model to find k).

Conclusion: To select roots from a given interval when solving a trigonometric equation, you need to:

1. to solve an equation of the form sin x = a, cos x = a It is more convenient to write the roots of the equation as two series of roots.
2. to solve equations of the form tan x = a, ctg x = a write down the general formula for roots.
3. create a mathematical model for each solution in the form of a double inequality and find the integer value of the parameter k or n.
4. substitute these values ​​into the root formula and calculate them.

Solve example No. 2 and No. 3 from homework using the resulting algorithm. Two students work at the board at the same time, followed by checking the work.

In this article I will try to explain 2 ways selecting roots in a trigonometric equation: using inequalities and using the trigonometric circle. Let's move straight to an illustrative example and we'll figure out how things work.

A) Solve the equation sqrt(2)cos^2x=sin(Pi/2+x)
b) Find all the roots of this equation belonging to the interval [-7Pi/2; -2Pi]

Let's solve point a.

Let's use the reduction formula for sine sin(Pi/2+x) = cos(x)

Sqrt(2)cos^2x = cosx

Sqrt(2)cos^2x - cosx = 0

Cosx(sqrt(2)cosx - 1) = 0

X1 = Pi/2 + Pin, n ∈ Z

Sqrt(2)cosx - 1 = 0

Cosx = 1/sqrt(2)

Cosx = sqrt(2)/2

X2 = arccos(sqrt(2)/2) + 2Pin, n ∈ Z
x3 = -arccos(sqrt(2)/2) + 2Pin, n ∈ Z

X2 = Pi/4 + 2Pin, n ∈ Z
x3 = -Pi/4 + 2Pin, n ∈ Z

Let's solve point b.

1) Selection of roots using inequalities

Here everything is done simply, we substitute the resulting roots into the interval given to us [-7Pi/2; -2Pi], find integer values ​​for n.

7Pi/2 less than or equal to Pi/2 + Pin less than or equal to -2Pi

We immediately divide everything by Pi

7/2 less than or equal to 1/2 + n less than or equal to -2

7/2 - 1/2 less than or equal to n less than or equal to -2 - 1/2

4 less than or equal to n less than or equal to -5/2

The integer n in this interval are -4 and -3. This means that the roots belonging to this interval will be Pi/2 + Pi(-4) = -7Pi/2, Pi/2 + Pi(-3) = -5Pi/2

Similarly we make two more inequalities

7Pi/2 less than or equal to Pi/4 + 2Pin less than or equal to -2Pi
-15/8 less than or equal to n less than or equal to -9/8

There are no whole n in this interval

7Pi/2 less than or equal to -Pi/4 + 2Pin less than or equal to -2Pi
-13/8 less than or equal to n less than or equal to -7/8

One integer n in this interval is -1. This means that the selected root on this interval is -Pi/4 + 2Pi*(-1) = -9Pi/4.

So the answer in point b: -7Pi/2, -5Pi/2, -9Pi/4

2) Selection of roots using a trigonometric circle

To use this method you need to understand how this circle works. I will try to explain in simple language how I understand this. I think in schools, during algebra lessons, this topic was explained many times with clever words from the teacher, in textbooks there were complex formulations. Personally, I understand this as a circle that can be walked around an infinite number of times, this is explained by the fact that the sine and cosine functions are periodic.

Let's go around counterclockwise

Let's go around 2 times counterclockwise

Let's go around 1 time clockwise (the values ​​will be negative)

Let's return to our question, we need to select roots in the interval [-7Pi/2; -2Pi]

To get to the numbers -7Pi/2 and -2Pi you need to go around the circle counterclockwise twice. In order to find the roots of the equation on this interval, you need to estimate and substitute.

Consider x = Pi/2 + Pin. Approximately what should n be for x to be somewhere in this range? We substitute, let's say -2, we get Pi/2 - 2Pi = -3Pi/2, obviously this is not included in our interval, so we take less than -3, Pi/2 - 3Pi = -5Pi/2, this is suitable, let's try again -4 , Pi/2 - 4Pi = -7Pi/2, also suitable.

Reasoning similarly for Pi/4 + 2Pin and -Pi/4 + 2Pin, we find another root -9Pi/4.

Comparison of two methods.

The first method (using inequalities) is much more reliable and much easier to understand, but if you really get serious about the trigonometric circle and the second selection method, then selecting roots will be much faster, you can save about 15 minutes on the exam.