Trigonometric formulas how to solve. Trigonometric Equations - Formulas, Solutions, Examples

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An equality containing an unknown under the sign of a trigonometric function (`sin x, cos x, tg x` or `ctg x`) is called a trigonometric equation, and we will consider their formulas 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's write the root formulas for each of them.

1. Equation `sin x=a`.

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

With `|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 the sine, there are no solutions among real numbers.

With `|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`

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

Methods for solving trigonometric equations

The solution of any trigonometric equation consists of two stages:

• using to convert it to the simplest;
• solve the resulting simple equation using the above formulas for the roots and tables.

Let's consider the main methods of solution using examples.

algebraic method.

In this method, the replacement of a variable and its substitution into equality is done.

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. Move to the left all terms of equality: `sin x+cos x-1=0`. Using , we transform and factorize the left 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 bring 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 split both parts by `cos x \ne 0` for the first case, and by `cos^2 x \ne 0` for the second. We get equations for `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, which must 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, dividing its left and right parts 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`, as a result `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. Applying the double angle formulas, the result is: `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 an auxiliary angle

In the trigonometric equation `a sin x + b cos x =c`, where a,b,c are coefficients and x is a variable, we divide both parts 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 modulus is not greater than 1. 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. Dividing both sides of the equation 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`.

Denote `3/5 = cos \varphi` , `4/5=sin \varphi`. Since `sin \varphi>0`, `cos \varphi>0`, 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, in the numerators and denominators of which there are trigonometric functions.

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

Solution. Multiply and divide the right side of the equation 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`

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

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. The study begins in the 10th grade, there are always tasks for the exam, so try to remember all the formulas of trigonometric equations - they will definitely come in handy for you!

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

When solving many math problems, especially those that occur before grade 10, the order of actions performed that will lead to the goal is clearly defined. Such problems include, for example, linear and quadratic equations, linear and quadratic inequalities, fractional equations and equations that reduce to quadratic. The principle of successful solution of each of the mentioned tasks is as follows: it is necessary to establish what type the problem being solved belongs to, remember the necessary sequence of actions that will lead to the desired result, i.e. answer and follow these steps.

Obviously, success or failure in solving a particular problem depends mainly on how correctly the type of the equation being solved is determined, how correctly the sequence of all stages of its solution is reproduced. Of course, in this case, it is necessary to have the skills to perform identical transformations and calculations.

A different situation occurs with trigonometric equations. It is not difficult to establish the fact that the equation is trigonometric. Difficulties arise when determining the sequence of actions that would lead to the correct answer.

It is sometimes difficult to determine its type by the appearance of an equation. And without knowing the type of equation, it is almost impossible to choose the right one from several dozen trigonometric formulas.

To solve the trigonometric equation, we must try:

1. bring all the functions included in the equation to "the same angles";
2. bring the equation to "the same functions";
3. factorize the left side of the equation, etc.

Consider basic methods for solving trigonometric equations.

I. Reduction to the simplest trigonometric equations

Solution scheme

Step 1. Express the trigonometric function in terms of known components.

Step 2 Find function argument using formulas:

cos x = a; x = ±arccos a + 2πn, n ЄZ.

sin x = a; x \u003d (-1) n arcsin a + πn, n Є Z.

tan x = a; x \u003d arctg a + πn, n Є Z.

ctg x = a; x \u003d arcctg a + πn, n Є Z.

Step 3 Find an unknown variable.

Example.

2 cos(3x – π/4) = -√2.

Solution.

1) cos(3x - π/4) = -√2/2.

2) 3x – π/4 = ±(π – π/4) + 2πn, n Є Z;

3x – π/4 = ±3π/4 + 2πn, n Є Z.

3) 3x = ±3π/4 + π/4 + 2πn, n Є Z;

x = ±3π/12 + π/12 + 2πn/3, n Є Z;

x = ±π/4 + π/12 + 2πn/3, n Є Z.

Answer: ±π/4 + π/12 + 2πn/3, n Є Z.

II. Variable substitution

Solution scheme

Step 1. Bring the equation to an algebraic form with respect to one of the trigonometric functions.

Step 2 Denote the resulting function by the variable t (if necessary, introduce restrictions on t).

Step 3 Write down and solve the resulting algebraic equation.

Step 4 Make a reverse substitution.

Step 5 Solve the simplest trigonometric equation.

Example.

2cos 2 (x/2) - 5sin (x/2) - 5 = 0.

Solution.

1) 2(1 - sin 2 (x/2)) - 5sin (x/2) - 5 = 0;

2sin 2(x/2) + 5sin(x/2) + 3 = 0.

2) Let sin (x/2) = t, where |t| ≤ 1.

3) 2t 2 + 5t + 3 = 0;

t = 1 or e = -3/2 does not satisfy the condition |t| ≤ 1.

4) sin (x/2) = 1.

5) x/2 = π/2 + 2πn, n Є Z;

x = π + 4πn, n Є Z.

Answer: x = π + 4πn, n Є Z.

III. Equation order reduction method

Solution scheme

Step 1. Replace this equation with a linear one using the power reduction formulas:

sin 2 x \u003d 1/2 (1 - cos 2x);

cos 2 x = 1/2 (1 + cos 2x);

tan 2 x = (1 - cos 2x) / (1 + cos 2x).

Step 2 Solve the resulting equation using methods I and II.

Example.

cos2x + cos2x = 5/4.

Solution.

1) cos 2x + 1/2 (1 + cos 2x) = 5/4.

2) cos 2x + 1/2 + 1/2 cos 2x = 5/4;

3/2 cos 2x = 3/4;

2x = ±π/3 + 2πn, n Є Z;

x = ±π/6 + πn, n Є Z.

Answer: x = ±π/6 + πn, n Є Z.

IV. Homogeneous equations

Solution scheme

Step 1. Bring this equation to the form

a) a sin x + b cos x = 0 (homogeneous equation of the first degree)

or to the view

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

Step 2 Divide both sides of the equation by

a) cos x ≠ 0;

b) cos 2 x ≠ 0;

and get the equation for tg x:

a) a tg x + b = 0;

b) a tg 2 x + b arctg x + c = 0.

Step 3 Solve the equation using known methods.

Example.

5sin 2 x + 3sin x cos x - 4 = 0.

Solution.

1) 5sin 2 x + 3sin x cos x – 4(sin 2 x + cos 2 x) = 0;

5sin 2 x + 3sin x cos x – 4sin² x – 4cos 2 x = 0;

sin 2 x + 3sin x cos x - 4cos 2 x \u003d 0 / cos 2 x ≠ 0.

2) tg 2 x + 3tg x - 4 = 0.

3) Let tg x = t, then

t 2 + 3t - 4 = 0;

t = 1 or t = -4, so

tg x = 1 or tg x = -4.

From the first equation x = π/4 + πn, n Є Z; from the second equation x = -arctg 4 + πk, k Є Z.

Answer: x = π/4 + πn, n Є Z; x \u003d -arctg 4 + πk, k Є Z.

V. Method for transforming an equation using trigonometric formulas

Solution scheme

Step 1. Using all kinds of trigonometric formulas, bring this equation to an equation that can be solved by methods I, II, III, IV.

Step 2 Solve the resulting equation using known methods.

Example.

sinx + sin2x + sin3x = 0.

Solution.

1) (sin x + sin 3x) + sin 2x = 0;

2sin 2x cos x + sin 2x = 0.

2) sin 2x (2cos x + 1) = 0;

sin 2x = 0 or 2cos x + 1 = 0;

From the first equation 2x = π/2 + πn, n Є Z; from the second equation cos x = -1/2.

We have x = π/4 + πn/2, n Є Z; from the second equation x = ±(π – π/3) + 2πk, k Є Z.

As a result, x \u003d π / 4 + πn / 2, n Є Z; x = ±2π/3 + 2πk, k Є Z.

Answer: x \u003d π / 4 + πn / 2, n Є Z; x = ±2π/3 + 2πk, k Є Z.

The ability and skills to solve trigonometric equations are very important, their development requires considerable effort, both on the part of the student and the teacher.

Many problems of stereometry, physics, etc. are associated with the solution of trigonometric equations. The process of solving such problems, as it were, contains many of the knowledge and skills that are acquired when studying the elements of trigonometry.

Trigonometric equations occupy an important place in the process of teaching mathematics and personality development in general.

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Trigonometry, as a science, originated in the Ancient East. The first trigonometric ratios were developed by astronomers to create an accurate calendar and orientate by the stars. These calculations were related to spherical trigonometry, while in school course study the ratio of the sides and angle of a flat triangle.

Trigonometry is a branch of mathematics dealing with the properties of trigonometric functions and the relationship between the sides and angles of triangles.

During the heyday of culture and science in the 1st millennium AD, knowledge spread from the Ancient East to Greece. But the main discoveries of trigonometry are the merit of the men of the Arab Caliphate. In particular, the Turkmen scientist al-Marazvi introduced such functions as tangent and cotangent, compiled the first tables of values ​​for sines, tangents and cotangents. The concept of sine and cosine was introduced by Indian scientists. A lot of attention is devoted to trigonometry in the works of such great figures of antiquity as Euclid, Archimedes and Eratosthenes.

Basic quantities of trigonometry

The basic trigonometric functions of a numerical argument are sine, cosine, tangent, and cotangent. Each of them has its own graph: sine, cosine, tangent and cotangent.

The formulas for calculating the values ​​of these quantities are based on the Pythagorean theorem. It is better known to schoolchildren in the formulation: “Pythagorean pants, equal in all directions,” since the proof is given on the example of an isosceles right triangle.

Sine, cosine and other dependencies establish a relationship between acute angles and sides of any right triangle. We give formulas for calculating these quantities for angle A and trace the relationship of trigonometric functions:

As you can see, tg and ctg are inverse functions. If we represent the leg a as the product of sin A and the hypotenuse c, and the leg b as cos A * c, then we get the following formulas for the tangent and cotangent:

trigonometric circle

Graphically, the ratio of the mentioned quantities can be represented as follows:

The circle, in this case, represents all possible values ​​of the angle α - from 0° to 360°. As can be seen from the figure, each function takes a negative or positive value depending on the angle. For example, sin α will be with a “+” sign if α belongs to the I and II quarters of the circle, that is, it is in the range from 0 ° to 180 °. With α from 180° to 360° (III and IV quarters), sin α can only be a negative value.

Let's try to build trigonometric tables for specific angles and find out the meaning of the quantities.

The values ​​of α equal to 30°, 45°, 60°, 90°, 180° and so on are called special cases. The values ​​of trigonometric functions for them are calculated and presented in the form of special tables.

These angles were not chosen by chance. The designation π in the tables is for radians. Rad is the angle at which the length of a circular arc corresponds to its radius. This value was introduced in order to establish a universal relationship; when calculating in radians, the actual length of the radius in cm does not matter.

The angles in the tables for trigonometric functions correspond to radian values:

So, it is not difficult to guess that 2π is a full circle or 360°.

Properties of trigonometric functions: sine and cosine

In order to consider and compare the basic properties of sine and cosine, tangent and cotangent, it is necessary to draw their functions. This can be done in the form of a curve located in a two-dimensional coordinate system.

Consider a comparative table of properties for a sine wave and a cosine wave:

sinusoid	cosine wave
y = sin x	y = cos x
ODZ [-1; 1]	ODZ [-1; 1]
sin x = 0, for x = πk, where k ϵ Z	cos x = 0, for x = π/2 + πk, where k ϵ Z
sin x = 1, for x = π/2 + 2πk, where k ϵ Z	cos x = 1, for x = 2πk, where k ϵ Z
sin x = - 1, at x = 3π/2 + 2πk, where k ϵ Z	cos x = - 1, for x = π + 2πk, where k ϵ Z
sin (-x) = - sin x, i.e. odd function	cos (-x) = cos x, i.e. the function is even
	the function is periodic, the smallest period is 2π
sin x › 0, with x belonging to quarters I and II or from 0° to 180° (2πk, π + 2πk)	cos x › 0, with x belonging to quarters I and IV or from 270° to 90° (- π/2 + 2πk, π/2 + 2πk)
sin x ‹ 0, with x belonging to quarters III and IV or from 180° to 360° (π + 2πk, 2π + 2πk)	cos x ‹ 0, with x belonging to quarters II and III or from 90° to 270° (π/2 + 2πk, 3π/2 + 2πk)
increases on the interval [- π/2 + 2πk, π/2 + 2πk]	increases on the interval [-π + 2πk, 2πk]
decreases on the intervals [ π/2 + 2πk, 3π/2 + 2πk]	decreases in intervals
derivative (sin x)' = cos x	derivative (cos x)’ = - sin x

Determining whether a function is even or not is very simple. It is enough to imagine a trigonometric circle with signs of trigonometric quantities and mentally “fold” the graph relative to the OX axis. If the signs are the same, the function is even; otherwise, it is odd.

The introduction of radians and the enumeration of the main properties of the sinusoid and cosine wave allow us to bring the following pattern:

It is very easy to verify the correctness of the formula. For example, for x = π/2, the sine is equal to 1, as is the cosine of x = 0. Checking can be done by looking at tables or by tracing function curves for given values.

Properties of tangentoid and cotangentoid

The graphs of the tangent and cotangent functions differ significantly from the sinusoid and cosine wave. The values ​​tg and ctg are inverse to each other.

1. Y = tgx.
2. The tangent tends to the values ​​of y at x = π/2 + πk, but never reaches them.
3. The smallest positive period of the tangentoid is π.
4. Tg (- x) \u003d - tg x, i.e., the function is odd.
5. Tg x = 0, for x = πk.
6. The function is increasing.
7. Tg x › 0, for x ϵ (πk, π/2 + πk).
8. Tg x ‹ 0, for x ϵ (— π/2 + πk, πk).
9. Derivative (tg x)' = 1/cos 2 ⁡x .

Consider the graphical representation of the cotangentoid below in the text.

The main properties of the cotangentoid:

1. Y = ctgx.
2. Unlike the sine and cosine functions, in the tangentoid Y can take on the values ​​of the set of all real numbers.
3. The cotangentoid tends to the values ​​of y at x = πk, but never reaches them.
4. The smallest positive period of the cotangentoid is π.
5. Ctg (- x) \u003d - ctg x, i.e., the function is odd.
6. Ctg x = 0, for x = π/2 + πk.
7. The function is decreasing.
8. Ctg x › 0, for x ϵ (πk, π/2 + πk).
9. Ctg x ‹ 0, for x ϵ (π/2 + πk, πk).
10. Derivative (ctg x)' = - 1/sin 2 ⁡x Fix

The concept of solving trigonometric equations.

• To solve a trigonometric equation, convert it to one or more basic trigonometric equations. Solving the trigonometric equation ultimately comes down to solving the four basic trigonometric equations.

Solution of basic trigonometric equations.

• There are 4 types of basic trigonometric equations:
• sin x = a; cos x = a
• tan x = a; ctg x = a
• Solving basic trigonometric equations involves looking at the various x positions on the unit circle, as well as using a conversion table (or calculator).
• Example 1. sin x = 0.866. Using a conversion table (or calculator), you get the answer: x = π/3. The unit circle gives another answer: 2π/3. Remember: all trigonometric functions are periodic, that is, their values ​​are repeated. For example, the periodicity of sin x and cos x is 2πn, and the periodicity of tg x and ctg x is πn. So the answer is written like this:
• x1 = π/3 + 2πn; x2 = 2π/3 + 2πn.
• Example 2 cos x = -1/2. Using a conversion table (or calculator), you get the answer: x = 2π/3. The unit circle gives another answer: -2π/3.
• x1 = 2π/3 + 2π; x2 = -2π/3 + 2π.
• Example 3. tg (x - π/4) = 0.
• Answer: x \u003d π / 4 + πn.
• Example 4. ctg 2x = 1.732.
• Answer: x \u003d π / 12 + πn.

Transformations used in solving trigonometric equations.

• To transform trigonometric equations, algebraic transformations (factoring, reduction of homogeneous terms, etc.) and trigonometric identities are used.
• Example 5. Using trigonometric identities, the equation sin x + sin 2x + sin 3x = 0 is converted to the equation 4cos x*sin (3x/2)*cos (x/2) = 0. Thus, the following basic trigonometric equations need to be solved: cos x = 0; sin(3x/2) = 0; cos(x/2) = 0.

• Finding angles from known values ​​of functions.

  • Before learning how to solve trigonometric equations, you need to learn how to find angles from known values ​​of functions. This can be done using a conversion table or calculator.
  • Example: cos x = 0.732. The calculator will give the answer x = 42.95 degrees. The unit circle will give additional angles, the cosine of which is also equal to 0.732.

• Set aside the solution on the unit circle.

  • You can put solutions to the trigonometric equation on the unit circle. The solutions of the trigonometric equation on the unit circle are the vertices of a regular polygon.
  • Example: The solutions x = π/3 + πn/2 on the unit circle are the vertices of the square.
  • Example: The solutions x = π/4 + πn/3 on the unit circle are the vertices of a regular hexagon.

• Methods for solving trigonometric equations.

  • If the given trigonometric equation contains only one trigonometric function, solve this equation as a basic trigonometric equation. If a given equation includes two or more trigonometric functions, then there are 2 methods for solving such an equation (depending on the possibility of its transformation).
    • Method 1
  • Transform this equation into an equation of the form: f(x)*g(x)*h(x) = 0, where f(x), g(x), h(x) are the basic trigonometric equations.
  • Example 6. 2cos x + sin 2x = 0. (0< x < 2π)
  • Solution. Using the double angle formula sin 2x = 2*sin x*cos x, replace sin 2x.
  • 2cos x + 2*sin x*cos x = 2cos x*(sin x + 1) = 0. Now solve two basic trigonometric equations: cos x = 0 and (sin x + 1) = 0.
  • Example 7 cos x + cos 2x + cos 3x = 0. (0< x < 2π)
  • Solution: Using trigonometric identities, transform this equation into an equation of the form: cos 2x(2cos x + 1) = 0. Now solve two basic trigonometric equations: cos 2x = 0 and (2cos x + 1) = 0.
  • Example 8. sin x - sin 3x \u003d cos 2x. (0< x < 2π)
  • Solution: Using trigonometric identities, transform this equation into an equation of the form: -cos 2x*(2sin x + 1) = 0. Now solve two basic trigonometric equations: cos 2x = 0 and (2sin x + 1) = 0.
    • Method 2
  • Convert the given trigonometric equation to an equation containing only one trigonometric function. Then replace this trigonometric function with some unknown, for example, t (sin x = t; cos x = t; cos 2x = t, tg x = t; tg (x/2) = t, etc.).
  • Example 9. 3sin^2 x - 2cos^2 x = 4sin x + 7 (0< x < 2π).
  • Solution. In this equation, replace (cos^2 x) with (1 - sin^2 x) (according to the identity). The transformed equation looks like:
  • 3sin^2 x - 2 + 2sin^2 x - 4sin x - 7 = 0. Replace sin x with t. Now the equation looks like: 5t^2 - 4t - 9 = 0. This is a quadratic equation with two roots: t1 = -1 and t2 = 9/5. The second root t2 does not satisfy the range of the function (-1< sin x < 1). Теперь решите: t = sin х = -1; х = 3π/2.
  • Example 10. tg x + 2 tg^2 x = ctg x + 2
  • Solution. Replace tg x with t. Rewrite the original equation as follows: (2t + 1)(t^2 - 1) = 0. Now find t and then find x for t = tg x.