Trigonometric functions how to solve examples. Trigonometric equations - formulas, solutions, examples

More complex trigonometric equations

Equations

sin x = a,
cos x = a,
tg x = a,
ctg x = a

are the simplest trigonometric equations. In this section, we will look at more complex trigonometric equations using specific examples. Their solution, as a rule, comes down to solving the simplest trigonometric equations.

Example 1 . Solve the equation

sin 2 X=cos X sin 2 x.

Transferring all terms of this equation to the left side and factoring the resulting expression, we obtain:

sin 2 X(1 - cos X) = 0.

The product of two expressions is equal to zero if and only if at least one of the factors is equal to zero, and the other takes any numerical value, as long as it is defined.

If sin 2 X = 0 , then 2 X= n π ; X = π / 2 n.

If 1 - cos X = 0 , then cos X = 1; X = 2kπ .

So, we got two groups of roots: X = π / 2 n; X = 2kπ . The second group of roots is obviously contained in the first, since for n = 4k the expression X = π / 2 n becomes
X = 2kπ .

Therefore, the answer can be written in one formula: X = π / 2 n, Where n- any integer.

Note that this equation could not be solved by reducing by sin 2 x. Indeed, after reduction we would get 1 - cos x = 0, whence X= 2k π . So we would lose some roots, for example π / 2 , π , 3π / 2 .

Example 2. Solve the equation

A fraction is equal to zero only if its numerator is equal to zero.
That's why sin 2 X = 0 , from where 2 X= n π ; X = π / 2 n.

From these values X you need to throw out as extraneous those values ​​at which sinX goes to zero (fractions with zero denominators have no meaning: division by zero is undefined). These values ​​are numbers that are multiples of π . In the formula
X = π / 2 n they are obtained for even n. Therefore, the roots of this equation will be the numbers

X = π / 2 (2k + 1),

where k is any integer.

Example 3 . Solve the equation

2 sin 2 X+ 7cos x - 5 = 0.

Let's express sin 2 X through cosx : sin 2 X = 1 - cos 2x . Then this equation can be rewritten as

2 (1 - cos 2 x) + 7cos x - 5 = 0 , or

2cos 2 x- 7 cos x + 3 = 0.

Designating cosx through at, we arrive at the quadratic equation

2у 2 - 7у + 3 = 0,

whose roots are the numbers 1/2 and 3. This means that either cos x= 1 / 2, or cos X= 3. However, the latter is impossible, since the cosine of any angle does not exceed 1 in absolute value.

It remains to admit that cos x = 1 / 2 , where

x = ± 60° + 360° n.

Example 4 . Solve the equation

2 sin X+ 3cos x = 6.

Since sin x and cos x in absolute value do not exceed 1, then the expression
2 sin X+ 3cos x cannot take values ​​greater than 5 . Therefore, this equation has no roots.

Example 5 . Solve the equation

sin X+cos x = 1

By squaring both sides of this equation, we get:

sin 2 X+ 2 sin x cos x+ cos 2 x = 1,

But sin 2 X + cos 2 x = 1 . That's why 2 sin x cos x = 0 . If sin x = 0 , That X = nπ ; if
cos x , That X = π / 2 + kπ . These two groups of solutions can be written in one formula:

X = π / 2 n

Since we squared both sides of this equation, it is possible that there are extraneous roots among the roots we obtained. That is why in this example, unlike all the previous ones, it is necessary to do a check. All meanings

X = π / 2 n can be divided into 4 groups

	1) X = 2kπ .	(n = 4k)
	2) X = π / 2 + 2kπ .	(n = 4k + 1)
	3) X = π + 2kπ .	(n = 4k + 2)
	4) X = 3π / 2 + 2kπ .	(n = 4k + 3)

At X = 2kπ sin x+cos x= 0 + 1 = 1. Therefore, X = 2kπ are the roots of this equation.

At X = π / 2 + 2kπ. sin x+cos x= 1 + 0 = 1 So X = π / 2 + 2kπ- also the roots of this equation.

At X = π + 2kπ sin x+cos x= 0 - 1 = - 1. Therefore, the values X = π + 2kπ are not roots of this equation. Similarly it is shown that X = 3π / 2 + 2kπ. are not roots.

Thus, this equation has the following roots: X = 2kπ And X = π / 2 + 2mπ., Where k And m- any integers.

When solving many mathematical 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 ones. The principle of successfully solving each of the mentioned problems is as follows: you need to establish what type of problem you are solving, remember the necessary sequence of actions that will lead to the desired result, i.e. answer and follow these steps.

It is obvious that success or failure in solving a particular problem depends mainly on how correctly the type of 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.

The situation is different with trigonometric equations. It is not at all 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 based on 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 a trigonometric equation, you need to try:

1. bring all functions included in the equation to “the same angles”;
2. bring the equation to “identical functions”;
3. factor the left side of the equation, etc.

Let's consider basic methods for solving trigonometric equations.

I. Reduction to the simplest trigonometric equations

Solution diagram

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

Step 2. Find the function argument using the formulas:

cos x = a; x = ±arccos a + 2πn, n ЄZ.

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

tan x = a; x = arctan a + πn, n Є Z.

ctg x = a; x = arcctg a + πn, n Є Z.

Step 3. Find the 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 replacement

Solution diagram

Step 1. Reduce the equation to 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 replacement.

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 diagram

Step 1. Replace this equation with a linear one, using the formula for reducing the degree:

sin 2 x = 1/2 · (1 – cos 2x);

cos 2 x = 1/2 · (1 + cos 2x);

tg 2 x = (1 – cos 2x) / (1 + cos 2x).

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

Example.

cos 2x + cos 2 x = 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 diagram

Step 1. Reduce 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 tan x:

a) a tan x + b = 0;

b) a tan 2 x + b arctan 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 = 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, which means

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 = -arctg 4 + πk, k Є Z.

V. Method of transforming an equation using trigonometric formulas

Solution diagram

Step 1. Using all possible trigonometric formulas, reduce this equation to an equation solved by methods I, II, III, IV.

Step 2. Solve the resulting equation using known methods.

Example.

sin x + sin 2x + sin 3x = 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 = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.

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

The ability and skill to solve trigonometric equations is very important, their development requires significant effort, both on the part of the student and on the part of the teacher.

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

Trigonometric equations occupy an important place in the process of learning mathematics and personal development in general.

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Requires knowledge of the basic formulas of trigonometry - the sum of the squares of sine and cosine, the expression of tangent through sine and cosine, and others. For those who have forgotten them or do not know them, we recommend reading the article "".
So, we know the basic trigonometric formulas, it's time to use them in practice. Solving trigonometric equations with the right approach, it’s quite an exciting activity, like, for example, solving a Rubik’s cube.

Based on the name itself, it is clear that a trigonometric equation is an equation in which the unknown is under the sign of the trigonometric function.
There are so-called simplest trigonometric equations. Here's what they look like: sinx = a, cos x = a, tan x = a. Let's consider how to solve such trigonometric equations, for clarity we will use the already familiar trigonometric circle.

sinx = a

cos x = a

tan x = a

cot x = a

Any trigonometric equation is solved in two stages: we reduce the equation to its simplest form and then solve it as a simple trigonometric equation.
There are 7 main methods by which trigonometric equations are solved.

1. Variable substitution and substitution method

Solve the equation 2cos 2 (x + /6) – 3sin( /3 – x) +1 = 0

Using the reduction formulas we get:

2cos 2 (x + /6) – 3cos(x + /6) +1 = 0

Replace cos(x + /6) with y to simplify and get the usual quadratic equation:

2y 2 – 3y + 1 + 0

The roots of which are y 1 = 1, y 2 = 1/2

Now let's go in reverse order

We substitute the found values ​​of y and get two answer options:

2. Solving trigonometric equations through factorization

How to solve the equation sin x + cos x = 1?

Let's move everything to the left so that 0 remains on the right:

sin x + cos x – 1 = 0

Let us use the identities discussed above to simplify the equation:

sin x - 2 sin 2 (x/2) = 0

Let's factorize:

2sin(x/2) * cos(x/2) - 2 sin 2 (x/2) = 0

2sin(x/2) * = 0

We get two equations

3. Reduction to a homogeneous equation

An equation is homogeneous with respect to sine and cosine if all its terms are relative to the sine and cosine of the same power of the same angle. To solve a homogeneous equation, proceed as follows:

a) transfer all its members to the left side;

b) take all common factors out of brackets;

c) equate all factors and brackets to 0;

d) a homogeneous equation of a lower degree is obtained in brackets, which in turn is divided into a sine or cosine of a higher degree;

e) solve the resulting equation for tg.

Solve the equation 3sin 2 x + 4 sin x cos x + 5 cos 2 x = 2

Let's use the formula sin 2 x + cos 2 x = 1 and get rid of the open two on the right:

3sin 2 x + 4 sin x cos x + 5 cos x = 2sin 2 x + 2cos 2 x

sin 2 x + 4 sin x cos x + 3 cos 2 x = 0

Divide by cos x:

tg 2 x + 4 tg x + 3 = 0

Replace tan x with y and get a quadratic equation:

y 2 + 4y +3 = 0, whose roots are y 1 =1, y 2 = 3

From here we find two solutions to the original equation:

x 2 = arctan 3 + k

4. Solving equations through the transition to a half angle

Solve the equation 3sin x – 5cos x = 7

Let's move on to x/2:

6sin(x/2) * cos(x/2) – 5cos 2 (x/2) + 5sin 2 (x/2) = 7sin 2 (x/2) + 7cos 2 (x/2)

Let's move everything to the left:

2sin 2 (x/2) – 6sin(x/2) * cos(x/2) + 12cos 2 (x/2) = 0

Divide by cos(x/2):

tg 2 (x/2) – 3tg(x/2) + 6 = 0

5. Introduction of auxiliary angle

For consideration, let’s take an equation of the form: a sin x + b cos x = c,

where a, b, c are some arbitrary coefficients, and x is an unknown.

Let's divide both sides of the equation by:



Now the coefficients of the equation, according to trigonometric formulas, have the properties sin and cos, namely: their modulus is not more than 1 and the sum of squares = 1. Let us denote them respectively as cos and sin, where - this is the so-called auxiliary angle. Then the equation will take the form:

cos * sin x + sin * cos x = C

or sin(x + ) = C

The solution to this simplest trigonometric equation is

x = (-1) k * arcsin C - + k, where



It should be noted that the notations cos and sin are interchangeable.

Solve the equation sin 3x – cos 3x = 1

The coefficients in this equation are:

a = , b = -1, so divide both sides by = 2

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 checking 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... 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.

When solving many mathematical 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 ones. The principle of successfully solving each of the mentioned problems is as follows: you need to establish what type of problem you are solving, remember the necessary sequence of actions that will lead to the desired result, i.e. answer and follow these steps.

It is obvious that success or failure in solving a particular problem depends mainly on how correctly the type of 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.

The situation is different with trigonometric equations. It is not at all 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 based on 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 a trigonometric equation, you need to try:

1. bring all functions included in the equation to “the same angles”;
2. bring the equation to “identical functions”;
3. factor the left side of the equation, etc.

Let's consider basic methods for solving trigonometric equations.

I. Reduction to the simplest trigonometric equations

Solution diagram

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

Step 2. Find the function argument using the formulas:

cos x = a; x = ±arccos a + 2πn, n ЄZ.

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

tan x = a; x = arctan a + πn, n Є Z.

ctg x = a; x = arcctg a + πn, n Є Z.

Step 3. Find the 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 replacement

Solution diagram

Step 1. Reduce the equation to 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 replacement.

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 diagram

Step 1. Replace this equation with a linear one, using the formula for reducing the degree:

sin 2 x = 1/2 · (1 – cos 2x);

cos 2 x = 1/2 · (1 + cos 2x);

tg 2 x = (1 – cos 2x) / (1 + cos 2x).

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

Example.

cos 2x + cos 2 x = 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 diagram

Step 1. Reduce 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 tan x:

a) a tan x + b = 0;

b) a tan 2 x + b arctan 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 = 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, which means

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 = -arctg 4 + πk, k Є Z.

V. Method of transforming an equation using trigonometric formulas

Solution diagram

Step 1. Using all possible trigonometric formulas, reduce this equation to an equation solved by methods I, II, III, IV.

Step 2. Solve the resulting equation using known methods.

Example.

sin x + sin 2x + sin 3x = 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 = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.

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

The ability and skill to solve trigonometric equations is very important, their development requires significant effort, both on the part of the student and on the part of the teacher.

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

Trigonometric equations occupy an important place in the process of learning mathematics and personal development in general.

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