a) Solve the equation: .

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

Problem solution

IN this lesson An example of solving a trigonometric equation is considered, 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 transformation is performed trigonometric function sine 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.

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

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

Show solution

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) By using 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] ;

Show solution

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].

Show solution

Solution

A) Because \sin \frac\pi 3=\cos \frac\pi 6, That \sin ^2\frac\pi 3=\cos ^2\frac\pi 6, Means, 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 quadratic equation relative to \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).

Show solution

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).

Note 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 arccosine 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). At the same time -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 ;

Show solution

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

is not included in the DZ.

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 side, 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 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 .

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.

You can order a detailed solution to your problem!!!

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 called `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`.

Go to half corner

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.

Mandatory minimum knowledge

sin x = a, -1 a 1 (a 1)
x = arcsin a + 2 n, n Z
x = - arcsin a + 2 n, n Z
or
x = (- 1)k arcsin a + k, k Z
arcsin (- a) = - arcsin a
sin x = 1
x = /2 + 2 k, k Z
sin x = 0
x = k, k Z
sin x = - 1
x = - /2 + 2 k, k Z
y
y
x
y
x
x

Mandatory minimum knowledge

cos x = a, -1 a 1 (a 1)
x = arccos a + 2 n, n Z
arccos (- a) = - arccos a
cos x = 1
x = 2 k, k Z
cos x = 0
x = /2 + k, k Z
y
y
x
cos x = - 1
x = + 2 k, k Z
y
x
x

Mandatory minimum knowledge

tg x = a, a R
x = arctan a + n, n Z
cot x = a, a R
x = arcctg a + n, n Z
arctg (- a) = - arctg a
arctg (- a) = - arctg a Reduce the equation to one function
Reduce to one argument
Some solution methods
trigonometric equations
Application of trigonometric formulas
Using abbreviated multiplication formulas
Factorization
Reduction to a quadratic equation for sin x, cos x, tan x
By introducing an auxiliary argument
By dividing both sides of a homogeneous equation of the first degree
(asin x +bcosx = 0) by cos x
By dividing both sides of a homogeneous equation of the second degree
(a sin2 x +bsin x cos x+ c cos2x =0) by cos2 x

Oral Exercises Calculate

arcsin ½
arcsin (- √2/2)
arccos √3/2
arccos (-1/2)
arctan √3
arctan (-√3/3)
= /6
= - /4
= /6
= - arccos ½ = - /3 = 2 /3
= /3
= - /6

(using a trigonometric circle)
cos 2x = ½, x [- /2; 3 /2]
2x = ± arccos ½ + 2 n, n Z
2x = ± /3 + 2 n, n Z
x = ± /6 + n, n Z
Let's select roots using a trigonometric circle
Answer: - /6; /6; 5 /6; 7 /6

Various methods of root selection

Find the roots of the equation belonging to the given interval
sin 3x = √3/2, x [- /2; /2]
3x = (– 1)k /3 + k, k Z
x = (– 1)k /9 + k/3, k Z
Let's select the roots by enumerating the values ​​of k:
k = 0, x = /9 – belongs to the interval
k = 1, x = – /9 + /3 = 2 /9 – belongs to the interval
k = 2, x = /9 + 2 /3 = 7 /9 – does not belong to the interval
k = – 1, x = – /9 – /3 = – 4 /9 – belongs to the interval
k = – 2, x = /9 – 2 /3 = – 5 /9 – does not belong to the interval
Answer: -4 /9; /9; 2 /9

Various methods of root selection

Find the roots of the equation belonging to the given interval
(using inequality)
tg 3x = – 1, x (- /2;)
3x = – /4 + n, n Z
x = – /12 + n/3, n Z
Let's select the roots using the inequality:
– /2 < – /12 + n/3 < ,
– 1/2 < – 1/12 + n/3 < 1,
– 1/2 + 1/12 < n/3 < 1+ 1/12,
– 5/12 < n/3 < 13/12,
– 5/4 < n < 13/4, n Z,
n = – 1; 0; 1; 2; 3
n = – 1, x = – /12 – /3 = – 5 /12
n = 0, x = – /12
n = 1, x = – /12 + /3 = /4
n = 2, x = – /12 + 2 /3 = 7 /12
n = 3, x = – /12 + = 11 /12
Answer: – 5 /12; – /12; /4; 7 /12; 11/12

10. Various methods of root selection

Find the roots of the equation belonging to the given interval
(using graph)
cos x = – √2/2, x [–4; 5 /4]
x = arccos (– √2/2) + 2 n, n Z
x = 3 /4 + 2 n, n Z
Let's select the roots using the graph:
x = – /2 – /4 = – 3 /4; x = – – /4 = – 5 /4
Answer: 5 /4; 3/4

11. 1. Solve the equation 72cosx = 49sin2x and indicate its roots on the segment [; 5/2]

1. Solve the equation 72cosx = 49sin2x
and indicate its roots on the segment [; 5 /2]
Let's solve the equation:
72cosx = 49sin2x,
72cosx = 72sin2x,
2cos x = 2sin 2x,
cos x – 2 sinx cosx = 0,
cos x (1 – 2sinx) = 0,
cos x = 0 ,
x = /2 + k, k Z
or
1 – 2sinx = 0,
sin x = ½,
x = (-1)n /6 + n, n Z
Let's select roots using
trigonometric circle:
x = 2 + /6 = 13 /6
Answer:
a) /2 + k, k Z, (-1)n /6 + n, n Z
b) 3 /2; 5 /2; 13/6

12. 2. Solve the equation 4cos2 x + 8 cos (x – 3/2) +1 = 0 Find its roots on the segment

2. Solve the equation 4cos2 x + 8 cos (x – 3 /2) +1 = 0
Find its roots on the segment
4cos2 x + 8 cos (x – 3 /2) +1 = 0
4cos2x + 8 cos (3 /2 – x) +1 = 0,
4cos2x – 8 sin x +1 = 0,
4 – 4sin2 x – 8 sin x +1 = 0,
4sin 2x + 8sin x – 5 = 0,
D/4 = 16 + 20 = 36,
sin x = – 2.5
or
sin x = ½
x = (-1)k /6 + k, k Z

13. Let’s select roots on a segment (using graphs)

Let's select roots on a segment
(using graphs)
sin x = ½
Let's plot the functions y = sin x and y = ½
x = 4 + /6 = 25 /6
Answer: a) (-1)k /6 + k, k Z; b) 25 /6

14. 3. Solve the equation Find its roots on the segment

4 – cos2 2x = 3 sin2 2x + 2 sin 4x
4 (sin2 2x + cos2 2x) – cos2 2x = 3 sin2 2x + 4 sin 2x cos 2x,
sin2 2x + 3 cos2 2x – 4 sin 2x cos 2x = 0
If cos2 2x = 0, then sin2 2x = 0, which is impossible, so
cos2 2x 0 and both sides of the equation can be divided by cos2 2x.
tg22x + 3 – 4 tg 2x = 0,
tg22x – 4 tg 2x + 3= 0,
tan 2x = 1,
2x = /4 + n, n Z
x = /8 + n/2, n Z
or
tan 2x = 3,
2x = arctan 3 + k, k Z
x = ½ arctan 3 + k/2, k Z

15.

4 – cos2 2x = 3 sin2 2x + 2 sin 4x
x = /8 + n/2, n Z or x = ½ arctan 3 + k/2, k Z
Since 0< arctg 3< /2,
0 < ½ arctg 3< /4, то ½ arctg 3
is the solution
Since 0< /8 < /4 < 1,значит /8
is also a solution
Other solutions will not be included in
gap since they
are obtained from the numbers ½ arctan 3 and /8
adding numbers that are multiples of /2.
Answer: a) /8 + n/2, n Z ; ½ arctan 3 + k/2, k Z
b) /8; ½ arctan 3

16. 4. Solve the equation log5(cos x – sin 2x + 25) = 2 Find its roots on the segment

4. Solve the equation log5(cos x – sin 2x + 25) = 2
Find its roots on the segment
Let's solve the equation:
log5(cos x – sin 2x + 25) = 2
ODZ: cos x – sin 2x + 25 > 0,
cos x – sin 2x + 25 = 25, 25 > 0,
cos x – 2sin x cos x = 0,
cos x (1 – 2sin x) = 0,
cos x = 0,
x = /2 + n, n Z
or
1 – 2sinx = 0,
sin x = 1/2
x = (-1)k /6 + k, k Z

17.

Let's select roots on a segment
Let's select roots on the segment:
1) x = /2 + n, n Z
2 /2 + n 7 /2, n Z
2 1/2 + n 7/2, n Z
2 – ½ n 7/2 – ½, n Z
1.5 n 3, n Z
n = 2; 3
x = /2 + 2 = 5 /2
x = /2 + 3 = 7 /2
2) sin x = 1/2
x = 2 + /6 = 13 /6
x = 3 – /6 = 17 /6
Answer: a) /2 + n, n Z ; (-1)k /6 + k, k Z
b) 13/6; 5 /2; 7 /2; 17/6

18. 5. Solve the equation 1/sin2x + 1/sin x = 2 Find its roots on the segment [-5/2; -3/2]

5. Solve the equation 1/sin2x + 1/sin x = 2
Find its roots on the segment [-5 /2; -3 /2]
Let's solve the equation:
1/sin2x + 1/sin x = 2
x k
Replacement 1/sin x = t,
t2 + t = 2,
t2 + t – 2 = 0,
t1= – 2, t2 = 1
1/sin x = – 2,
sin x = – ½,
x = – /6 + 2 n, n Z
or
x = – 5 /6 + 2 n, n Z
1/sin x = 1,
sin x = 1,
x = /2 + 2 n, n Z
This series of roots is excluded, because -150º+ 360ºn is outside the limits
specified interval [-450º; -270º]

19.

Let's continue selecting roots on the segment
Let's consider the remaining series of roots and carry out a selection of roots
on the segment [-5 /2; -3 /2] ([-450º; -270º]):
1) x = - /6 + 2 n, n Z
2) x = /2 + 2 n, n Z
-5 /2 - /6 + 2 n -3 /2, n Z
-5 /2 /2 + 2 n -3 /2, n Z
-5/2 -1/6 + 2n -3/2, n Z
-5/2 1/2 + 2n -3/2, n Z
-5/2 +1/6 2n -3/2 + 1/6, n Z
-5/2 - 1/2 2n -3/2 - 1/2, n Z
– 7/3 2n -4/3, n Z
– 3 2n -2, n Z
-7/6 n -2/3, n Z
-1.5 n -1. n Z
n = -1
n = -1
x = - /6 - 2 = -13 /6 (-390º)
x = /2 - 2 = -3 /2 (-270º)
Answer: a) /2 + 2 n, n Z ; (-1)k+1 /6 + k, k Z
b) -13 /6; -3 /2

20. 6. Solve the equation |sin x|/sin x + 2 = 2cos x Find its roots on the segment [-1; 8]

Let's solve the equation
|sin x|/sin x + 2 = 2cos x
1)If sin x >0, then |sin x| =sin x
The equation will take the form:
2 cos x=3,
cos x =1.5 – has no roots
2) If sin x<0, то |sin x| =-sin x
and the equation will take the form
2cos x=1, cos x = 1/2,
x = ±π/3 +2πk, k Z
Considering that sin x< 0, то
one series of answers left
x = - π/3 +2πk, k Z
Let's select roots for
segment [-1; 8]
k=0, x= - π/3 , - π< -3, - π/3 < -1,
-π/3 does not belong to this
segment
k=1, x = - π/3 +2π = 5π/3<8,
5 π/3 [-1; 8]
k=2, x= - π/3 + 4π = 11π/3 > 8,
11π/3 does not belong to this
segment.
Answer: a) - π/3 +2πk, k Z
b) 5
π/3

21. 7. Solve the equation 4sin3x=3cos(x- π/2) Find its roots on the interval

8. Solve the equation √1-sin2x= sin x
Find its roots on the interval
Let's solve the equation √1-sin2x= sin x.
sin x ≥ 0,
1- sin2x = sin2x;
sin x ≥ 0,
2sin2x = 1;
sin x≥0,
sin x =√2/2; sin x = - √2/2;
sin x =√2/2
x=(-1)k /4 + k, k Z
sin x =√2/2

25. Let’s select roots on a segment

Let's select roots on a segment
x=(-1)k /4 + k, k Z
sin x =√2/2
y =sin x and y=√2/2
5 /2 + /4 = 11 /4
Answer: a) (-1)k /4 + k, k Z; b) 11 /4

26. 9. Solve the equation (sin2x + 2 sin2x)/√-cos x =0 Find its roots on the interval [-5; -7/2]

9. Solve the equation (sin2x + 2 sin2x)/√-cos x =0
Find its roots on the interval [-5; -7 /2]
Let's solve the equation
(sin2x + 2 sin2x)/√-cos x =0.
1) ODZ: cos x<0 ,
/2 +2 n 2) sin2x + 2 sin2x =0,
2 sinx∙cos x + 2 sin2x =0,
sin x (cos x+ sin x) =0,
sin x=0, x= n, n Z
or
cos x+ sin x=0 | : cos x,
tan x= -1, x= - /4 + n, n Z
Taking into account DL
x= n, n Z, x= +2 n, n Z;
x= - /4 + n, n Z,
x= 3 /4 + 2 n, n Z

27. Let's select roots on a given segment

Let's select roots on the given
segment [-5; -7 /2]
x= +2 n, n Z ;
-5 ≤ +2 n ≤ -7 /2,
-5-1 ≤ 2n ≤ -7/2-1,
-3≤ n ≤ -9/4, n Z
n = -3, x= -6 = -5
x= 3 /4 + 2 n, n Z
-5 ≤ 3 /4 + 2 n ≤ -7 /2
-23/8 ≤ n ≤ -17/8, no such thing
whole n.
Answer: a) +2 n, n Z ;
3 /4 + 2 n, n Z ;
b) -5.

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

10. Solve the equation 2sin2x =4cos x –sinx+1
Find its roots on the interval [ /2; 3 /2]
Let's solve the equation
2sin2x = 4cos x – sinx+1
2sin2x = 4cos x – sinx+1,
4 sinx∙cos x – 4cos x + sin x -1 = 0,
4cos x(sin x – 1) + (sin x – 1) = 0,
(sin x – 1)(4cos x +1)=0,
sin x – 1= 0, sin x = 1, x = /2+2 n, n Z
or
4cos x +1= 0, cos x = -0.25
x = ± (-arccos (0.25)) + 2 n, n Z
Let's write the roots of this equation differently
x = - arccos(0.25) + 2 n,
x = -(- arccos(0.25)) + 2 n, n Z

29. Let's select roots using a circle

x = /2+2 n, n Z, x = /2;
x = -arccos(0.25)+2 n,
x=-(-arccos(0.25)) +2 n, n Z,
x = - arccos(0.25),
x = + arccos(0.25)
Answer: a) /2+2 n,
-arccos(0.25)+2 n,
-(-arccos(0.25)) +2 n, n Z;
b) /2;
-arccos(0.25); +arccos(0.25)