Tangent equation formula. The simplest trigonometric equations. Protection of personal information

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The video course “Get an A” includes all the topics necessary to successfully pass the Unified State Exam in mathematics with 60-65 points. Completely all tasks 1-13 of the Profile Unified State Exam in mathematics. Also suitable for passing the Basic Unified State Examination in mathematics. If you want to pass the Unified State Exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!

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The simplest trigonometric equations are solved, as a rule, using formulas. Let me remind you that the simplest trigonometric equations are:

sinx = a

cosx = a

tgx = a

ctgx = a

x is the angle to be found,
a is any number.

And here are the formulas with which you can immediately write down the solutions to these simplest equations.

For sine:

For cosine:

x = ± arccos a + 2π n, n ∈ Z

For tangent:

x = arctan a + π n, n ∈ Z

For cotangent:

x = arcctg a + π n, n ∈ Z

Actually, this is the theoretical part of solving the simplest trigonometric equations. Moreover, everything!) Nothing at all. However, the number of errors on this topic is simply off the charts. Especially if the example deviates slightly from the template. Why?

Yes, because a lot of people write down these letters, without understanding their meaning at all! He writes down with caution, lest something happen...) This needs to be sorted out. Trigonometry for people, or people for trigonometry, after all!?)

Let's figure it out?

One angle will be equal to arccos a, second: -arccos a.

And it will always work out this way. For any A.

If you don’t believe me, hover your mouse over the picture, or touch the picture on your tablet.) I changed the number A to something negative. Anyway, we got one corner arccos a, second: -arccos a.

Therefore, the answer can always be written as two series of roots:

x 1 = arccos a + 2π n, n ∈ Z

x 2 = - arccos a + 2π n, n ∈ Z

Let's combine these two series into one:

x= ± arccos a + 2π n, n ∈ Z

And that's all. We have obtained a general formula for solving the simplest trigonometric equation with cosine.

If you understand that this is not some kind of superscientific wisdom, but just a shortened version of two series of answers, You will also be able to handle tasks “C”. With inequalities, with selecting roots from a given interval... There the answer with a plus/minus does not work. But if you treat the answer in a businesslike manner, and break it down into two separate answers, everything will be resolved.) Actually, that’s why we’re looking into it. What, how and where.

In the simplest trigonometric equation

sinx = a

we also get two series of roots. Always. And these two series can also be recorded in one line. Only this line will be trickier:

x = (-1) n arcsin a + π n, n ∈ Z

But the essence remains the same. Mathematicians simply designed a formula to make one instead of two entries for series of roots. That's all!

Let's check the mathematicians? And you never know...)

In the previous lesson, the solution (without any formulas) of a trigonometric equation with sine was discussed in detail:

The answer resulted in two series of roots:

x 1 = π /6 + 2π n, n ∈ Z

x 2 = 5π /6 + 2π n, n ∈ Z

If we solve the same equation using the formula, we get the answer:

x = (-1) n arcsin 0.5 + π n, n ∈ Z

Actually, this is an unfinished answer.) The student must know that arcsin 0.5 = π /6. The complete answer would be:

x = (-1)n π /6+ π n, n ∈ Z

This raises an interesting question. Reply via x 1; x 2 (this is the correct answer!) and through lonely X (and this is the correct answer!) - are they the same thing or not? We'll find out now.)

We substitute in the answer with x 1 values n =0; 1; 2; etc., we count, we get a series of roots:

x 1 = π/6; 13π/6; 25π/6 and so on.

With the same substitution in response with x 2 , we get:

x 2 = 5π/6; 17π/6; 29π/6 and so on.

Now let's substitute the values n (0; 1; 2; 3; 4...) into the general formula for single X . That is, we raise minus one to the zero power, then to the first, second, etc. Well, of course, we substitute 0 into the second term; 1; 2 3; 4, etc. And we count. We get the series:

x = π/6; 5π/6; 13π/6; 17π/6; 25π/6 and so on.

That's all you can see.) The general formula gives us exactly the same results as are the two answers separately. Just everything at once, in order. The mathematicians were not fooled.)

Formulas for solving trigonometric equations with tangent and cotangent can also be checked. But we won’t.) They are already simple.

I wrote out all this substitution and verification specifically. Here it is important to understand one simple thing: there are formulas for solving elementary trigonometric equations, just a short summary of the answers. For this brevity, we had to insert plus/minus into the cosine solution and (-1) n into the sine solution.

These inserts do not interfere in any way in tasks where you just need to write down the answer to an elementary equation. But if you need to solve an inequality, or then you need to do something with the answer: select roots on an interval, check for ODZ, etc., these insertions can easily unsettle a person.

So what should I do? Yes, either write the answer in two series, or solve the equation/inequality using the trigonometric circle. Then these insertions disappear and life becomes easier.)

We can summarize.

To solve the simplest trigonometric equations, there are ready-made answer formulas. Four pieces. They are good for instantly writing down the solution to an equation. For example, you need to solve the equations:

sinx = 0.3

Easily: x = (-1) n arcsin 0.3 + π n, n ∈ Z

cosx = 0.2

No problem: x = ± arccos 0.2 + 2π n, n ∈ Z

tgx = 1.2

Easily: x = arctan 1,2 + π n, n ∈ Z

ctgx = 3.7

One left: x= arcctg3,7 + π n, n ∈ Z

cos x = 1.8

If you, shining with knowledge, instantly write the answer:

x= ± arccos 1.8 + 2π n, n ∈ Z

then you are already shining, this is... that... from a puddle.) Correct answer: there are no solutions. Don't understand why? Read what arc cosine is. In addition, if on the right side of the original equation there are tabular values ​​of sine, cosine, tangent, cotangent, - 1; 0; √3; 1/2; √3/2 and so on. - the answer through the arches will be unfinished. Arches must be converted to radians.

And if you come across inequality, like

then the answer is:

x πn, n ∈ Z

there is rare nonsense, yes...) Here you need to solve using the trigonometric circle. What we will do in the corresponding topic.

For those who heroically read to these lines. I simply cannot help but appreciate your titanic efforts. Bonus for you.)

Bonus:

When writing down formulas in an alarming combat situation, even seasoned nerds often get confused about where πn, And where 2π n. Here's a simple trick for you. In everyone formulas worth πn. Except for the only formula with arc cosine. It stands there 2πn. Two peen. Keyword - two. In this same formula there are two sign at the beginning. Plus and minus. Here and there - two.

So if you wrote two sign before the arc cosine, it’s easier to remember what will happen at the end two peen. And it also happens the other way around. The person will miss the sign ± , gets to the end, writes correctly two Pien, and he’ll come to his senses. There's something ahead two sign! The person will return to the beginning and correct the mistake! Like this.)

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

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 + 2nn,

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).

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 `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.