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Solution of the equation cot x a. The simplest trigonometric equations

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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`,

`sin x/2 =0`, `x/2 =\pi n`, `x_1=2\pi n`.
`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:

`tg x=-2`, `x_1=arctg (-2)+\pi n`, `n \in Z`
`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:

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

`sin x=0`, `x=\pi n`, `n \in Z`
`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.

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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

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.

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 + 2nn,

Equation sin(x) = a