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 of the problem to be solved, to 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 the equation being solved is determined, how correctly the sequence of all stages of its solution is reproduced. Of course, it is necessary to have skills in performing identical transformations and calculations.
The situation is different with trigonometric equations. Establishing the fact that the equation is trigonometric is not difficult at all. Difficulties arise in determining the sequence of actions that would lead to the correct answer.
The appearance of an equation can sometimes be difficult to determine its type. And without knowing the type of equation, it is almost impossible to choose the desired one from several tens of trigonometric formulas.
To solve the trigonometric equation, one should try:
1. bring all the functions included in the equation to "equal angles";
2. to bring the equation to "the same functions";
3. factor 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 a trigonometric function in terms of known components.
Step 2. Find the argument of a function by the formulas:
cos x = a; x = ± arccos a + 2πn, n ЄZ.
sin x = a; x = (-1) n arcsin a + πn, n Є Z.
tg x = a; x = arctan a + πn, n Є Z.
ctg x = a; x = arcctg a + πn, n Є Z.
Step 3. Find 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 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 scheme
Step 1. Replace the given equation with a linear one, using the degree reduction formulas for this:
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 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 mind
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 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, 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 = -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 the equation solved by methods I, II, III, IV.
Step 2. Solve the resulting equation by 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 to solve trigonometric equations is very 
important, their development requires significant efforts, both on the part of the student and on the part of the teacher.
Many problems of stereometry, physics, etc. are connected with the solution of trigonometric equations. The process of solving such problems, as it were, contains many 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 the development of personality in general.
Still have questions? Not sure how to solve trigonometric equations?
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Trigonometric equations are not the easiest topic. Painfully they are diverse.) For example, such:
sin 2 x + cos3x = ctg5x
sin (5x + π / 4) = ctg (2x-π / 3)
sinx + cos2x + tg3x = ctg4x
Etc...
But these (and all others) trigonometric monsters have two common and obligatory characteristics. The first - you will not believe - there are trigonometric functions in the equations.) Second: all expressions with x are found inside these same functions. And only there! If x appears anywhere outside, for example, sin2x + 3x = 3, this will already be a mixed-type equation. Such equations require an individual approach. We will not consider them here.
We will not solve evil equations in this lesson either.) Here we will deal with the most simple trigonometric equations. Why? Yes, because the solution any trigonometric equations have two stages. At the first stage, the evil equation is reduced to a simple one by means of various transformations. On the second, this simplest equation is solved. No other way.
So, if you have problems at the second stage, the first stage does not make much sense.)
What do elementary trigonometric equations look like?
sinx = a
cosx = a
tgx = a
ctgx = a
Here a denotes any number. Anyone.
By the way, inside the function there may not be a pure x, but some kind of expression, such as:
cos (3x + π / 3) = 1/2
etc. This complicates life, but it does not affect the method of solving the trigonometric equation.
How to solve trigonometric equations?
Trigonometric equations can be solved in two ways. First way: using logic and trigonometric circle. We will consider this path here. The second way - using memory and formulas - will be discussed in the next lesson.
The first way is clear, reliable, and hard to forget.) It is good for solving trigonometric equations, inequalities, and all sorts of tricky non-standard examples. Logic is stronger than memory!)
Solving equations using the trigonometric circle.
We include elementary logic and the ability to use the trigonometric circle. Don't know how !? However ... It's difficult for you in trigonometry ...) But it doesn't matter. Take a look at the lessons "Trigonometric circle ...... What is it?" and "Counting angles on a trigonometric circle". Everything is simple there. Unlike tutorials ...)
Oh, you know !? And even mastered the "Practical work with the trigonometric circle" !? Congratulations. This topic will be close and understandable to you.) What is especially pleasing, the trigonometric circle does not care which equation you solve. Sine, cosine, tangent, cotangent - everything is one for him. There is only one solution principle.
So we take any elementary trigonometric equation. At least this:
cosx = 0.5
We need to find the X. In human terms, you need find the angle (x), the cosine of which is 0.5.
How did we use the circle earlier? We drew a corner on it. In degrees or radians. And immediately seen trigonometric functions of this angle. Now let's do the opposite. Let's draw a cosine equal to 0.5 on the circle and immediately see injection. All that remains is to write down the answer.) Yes, yes!
Draw a circle and mark a cosine of 0.5. On the cosine axis, of course. Like this:
Now let's draw the angle that this cosine gives us. Move the mouse cursor over the drawing (or tap the picture on the tablet), and see this very corner NS.
What angle is cosine 0.5?
x = π / 3
cos 60 °= cos ( π / 3) = 0,5
Someone will chuckle skeptically, yes ... They say, was it worth the circle, when everything is already clear ... You can, of course, chuckle ...) But the fact is that this is an erroneous answer. Or rather, insufficient. Connoisseurs of the circle understand that there are still a whole bunch of angles here, which also give a cosine equal to 0.5.
If you turn the movable side of the OA full turn, point A will return to its original position. With the same cosine equal to 0.5. Those. the angle will change 360 ° or 2π radians, and cosine is not. The new angle 60 ° + 360 ° = 420 ° will also be the solution to our equation, since
You can wind an infinite number of such full turns ... And all these new angles will be solutions to our trigonometric equation. And all of them must somehow be written down in response. Everything. Otherwise, the decision does not count, yes ...)
Mathematics knows how to do this in a simple and elegant way. In one short answer, write endless set solutions. This is what it looks like for our equation:
x = π / 3 + 2π n, n ∈ Z
I will decipher. Still write meaningfully more pleasant than stupidly drawing some mysterious letters, right?)
π / 3 - this is the same corner that we saw on the circle and identified according to the cosine table.
2π is one complete revolution in radians.
n is the number of full, i.e. whole revolutions. It is clear that n can be 0, ± 1, ± 2, ± 3 .... and so on. As indicated by a short note:
n ∈ Z
n belongs ( ∈ ) to the set of integers ( Z ). By the way, instead of the letter n letters may well be used k, m, t etc.
This entry means that you can take any whole n ... At least -3, at least 0, at least +55. What you want. If you plug that number into your answer, you get a specific angle that will definitely solve our harsh equation.)
Or, in other words, x = π / 3 is the only root of the infinite set. To get all the other roots, it is enough to add any number of full revolutions to π / 3 ( n ) in radians. Those. 2π n radian.
Everything? No. I deliberately stretch the pleasure. To remember it better.) We received only part of the answers to our equation. I will write this first part of the solution as follows:
x 1 = π / 3 + 2π n, n ∈ Z
x 1 - not one root, it is a whole series of roots, written in short form.
But there are also angles that also give a cosine of 0.5!
Let's go back to our picture, which was used to write down the answer. There she is:
Hover the mouse over the picture and see another corner that also gives a cosine of 0.5. What do you think it is equal to? The triangles are the same ... Yes! It is equal to the corner NS , only put back in the negative direction. This is the corner -NS. But we have already figured out the x. π / 3 or 60 °. Therefore, we can safely write:
x 2 = - π / 3
Well, of course, we add all the angles that are obtained through full revolutions:
x 2 = - π / 3 + 2π n, n ∈ Z
Now that's it.) In the trigonometric circle, we saw(who understands, of course)) all angles giving a cosine equal to 0.5. And they wrote these angles in short mathematical form. The answer produced two endless series of roots:
x 1 = π / 3 + 2π n, n ∈ Z
x 2 = - π / 3 + 2π n, n ∈ Z
This is the correct answer.
Hope, general principle of solving trigonometric equations using a circle is clear. We mark on the circle the cosine (sine, tangent, cotangent) from the given equation, draw the angles corresponding to it and write down the answer. Of course, you need to figure out what kind of corners we are saw on the circle. Sometimes it's not that obvious. Well, so I said that logic is required here.)
For example, let's look at another trigonometric equation:
Please note that the number 0.5 is not the only possible number in the equations!) It's just more convenient for me to write it than roots and fractions.
We work according to the general principle. Draw a circle, mark (on the sine axis, of course!) 0.5. We draw at once all the angles corresponding to this sine. Let's get the following picture:
Dealing with the angle first NS in the first quarter. We recall the table of sines and determine the value of this angle. It's a simple matter:
x = π / 6
We remember the full turns and, with a clear conscience, write down the first series of answers:
x 1 = π / 6 + 2π n, n ∈ Z
Half done. But now we need to define second corner ... This is more cunning than in cosines, yes ... But logic will save us! How to determine the second angle through x? Yes Easy! The triangles in the picture are the same, and the red corner NS equal to the angle NS ... Only it is measured from the angle π in the negative direction. Therefore, it is red.) And for the answer we need an angle, measured correctly, from the positive OX semiaxis, i.e. from an angle of 0 degrees.
Hover the cursor over the picture and see everything. I removed the first corner so as not to complicate the picture. The angle we are interested in (drawn in green) will be equal to:
π - x
X we know it π / 6 ... Therefore, the second angle will be:
π - π / 6 = 5π / 6
We again recall the addition of full revolutions and write down the second series of responses:
x 2 = 5π / 6 + 2π n, n ∈ Z
That's all. The complete answer consists of two series of roots:
x 1 = π / 6 + 2π n, n ∈ Z
x 2 = 5π / 6 + 2π n, n ∈ Z
Equations with tangent and cotangent can be easily solved using the same general principle for solving trigonometric equations. If, of course, you know how to draw tangent and cotangent on a trigonometric circle.
In the examples above, I used the table sine and cosine value: 0.5. Those. one of those meanings that the student knows must. Now let's expand our capabilities to all other values. Decide, so decide!)
So, let's say we need to solve this trigonometric equation:
There is no such cosine value in short tables. We ignore this terrible fact in cold blood. Draw a circle, mark 2/3 on the cosine axis and draw the corresponding angles. We get just such a picture.
Let's figure it out, for a start, with an angle in the first quarter. If I had known what the X was, they would have written down the answer right away! We don't know ... Failure !? Calm! Math does not abandon its own in trouble! She came up with arccosines for this case. Do not know? In vain. Find out, It's much easier than you think. Under this link, there is not a single tricky incantation about "inverse trigonometric functions" ... This is superfluous in this topic.
If you are in the know, it is enough to say to yourself: "X is the angle, the cosine of which is 2/3". And right away, purely by the definition of the arccosine, you can write:
We recall additional turns and calmly write down the first series of roots of our trigonometric equation:
x 1 = arccos 2/3 + 2π n, n ∈ Z
The second series of roots is also recorded almost automatically for the second angle. Everything is the same, only x (arccos 2/3) will be with a minus:
x 2 = - arccos 2/3 + 2π n, n ∈ Z
And that's all! This is the correct answer. Even easier than with table values. You don't need to remember anything.) By the way, the most attentive will notice that this picture with the solution through the inverse cosine in essence, does not differ from the picture for the equation cosx = 0.5.
Exactly! The general principle is the general one! I specially drew two almost identical pictures. The circle shows us the angle NS by its cosine. The table is a cosine, or not - the circle does not know. What is this angle, π / 3, or what kind of inverse cosine - that's up to us.
With sine the same song. For example:
Draw the circle again, mark the sine equal to 1/3, draw the corners. The picture looks like this:
And again the picture is almost the same as for the equation sinx = 0.5. Again, start at the corner in the first quarter. What is x if its sine is 1/3? No problem!
So the first pack of roots is ready:
x 1 = arcsin 1/3 + 2π n, n ∈ Z
We deal with the second corner. In the example with a table value of 0.5, it was:
π - x
So here it will be exactly the same! Only x is different, arcsin 1/3. So what!? You can safely write down the second pack of roots:
x 2 = π - arcsin 1/3 + 2π n, n ∈ Z
This is an absolutely correct answer. Although it doesn't look very familiar. But it is understandable, I hope.)
This is how trigonometric equations are solved using a circle. This path is clear and understandable. It is he who saves in trigonometric equations with the selection of roots at a given interval, in trigonometric inequalities - they are generally solved almost always in a circle. In short, in any tasks that are slightly more difficult than the standard ones.
Let's apply our knowledge in practice?)
Solve trigonometric equations:
At first it's simpler, right from this lesson.
Now more difficult.
Hint: This is where you have to reflect on the circle. Personally.)
And now they are outwardly unpretentious ... They are also called special cases.
sinx = 0
sinx = 1
cosx = 0
cosx = -1
Hint: here you need to figure out in a circle where there are two series of answers, and where is one ... And how to write down one instead of two series of answers. Yes, so that not a single root of the infinite number is lost!)
Well, very simple ones):
sinx = 0,3
cosx = π
tgx = 1,2
ctgx = 3,7
Hint: here you need to know what is arcsine, arccosine? What is arc tangent, arc cotangent? The simplest definitions. But you don't need to remember any table values!)
The answers are, of course, a mess):
x 1= arcsin0,3 + 2π n, n ∈ Z
x 2= π - arcsin0,3 + 2
Not everything works out? It happens. Read the lesson again. Only thoughtfully(there is such an outdated word ...) And follow the links. The main links are about the circle. Without it, in trigonometry, it's like crossing the road with a blindfold. Sometimes it works.)
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. Instant validation testing. Learning - with interest!)
you can get acquainted with functions and derivatives.
Requires knowledge of the basic formulas of trigonometry - the sum of the squares of the sine and cosine, the expression of the tangent through the sine and cosine, and others. For those who have forgotten them or do not know, 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 is 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 the so-called simplest trigonometric equations. This is how they look: sinx = a, cos x = a, tg x = a. Consider how to solve such trigonometric equations, for clarity, we will use the already familiar trigonometric circle.
sinx = a
cos x = a
tg x = a
cot x = a
Any trigonometric equation is solved in two stages: we bring the equation to the simplest form and then solve it as the simplest trigonometric equation.
There are 7 main methods by which trigonometric equations are solved.
Variable Substitution and Substitution Method
Solving trigonometric equations through factorization
Reduction to a homogeneous equation
Solving equations by going to half angle
Introduce an auxiliary angle
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 for simplicity and get the usual quadratic equation:
2y 2 - 3y + 1 + 0
Whose roots y 1 = 1, y 2 = 1/2
Now let's go in reverse order
We substitute the found y values and we get two answers:
How to solve the equation sin x + cos x = 1?
Move everything to the left so that 0 remains on the right:
sin x + cos x - 1 = 0
Let's use the above identities to simplify the equation:
sin x - 2 sin 2 (x / 2) = 0
We do the factorization:
2sin (x / 2) * cos (x / 2) - 2 sin 2 (x / 2) = 0
2sin (x / 2) * = 0
We get two equations
An equation is homogeneous with respect to sine and cosine if all of its terms with respect to sine and cosine are the same power of the same angle. To solve a homogeneous equation, proceed as follows:
a) transfer all of its members to the left side;
b) take all common factors out of parentheses;
c) equate all factors and brackets to 0;
d) a homogeneous equation of a lesser degree is obtained in brackets, it in turn is divided into sine or cosine in the highest 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 tg x with y and get a quadratic equation:
y 2 + 4y +3 = 0, whose roots y 1 = 1, y 2 = 3
From here we find two solutions to the original equation:
x 2 = arctan 3 + k
Solve the equation 3sin x - 5cos x = 7
Moving 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)
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
For consideration, take an equation of the form: a sin x + b cos x = c,
where a, b, c are some arbitrary coefficients, and x is unknown.
Divide both sides of the equation into:
Now the coefficients of the equation, according to the trigonometric formulas, have the properties of sin and cos, namely: their modulus is not more than 1 and the sum of squares = 1. Let's denote them as cos and sin, respectively, where is the so-called auxiliary angle. Then the equation will take the form:
cos * sin x + sin * cos x = С
or sin (x +) = C
The solution to this simplest trigonometric equation is
x = (-1) k * arcsin С - + k, where
Note that cos and sin are used interchangeably.
Solve the equation sin 3x - cos 3x = 1
In this equation, the coefficients are:
a =, b = -1, so we divide both sides by = 2
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 of the problem to be solved, to 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 the equation being solved is determined, how correctly the sequence of all stages of its solution is reproduced. Of course, it is necessary to have skills in performing identical transformations and calculations.
The situation is different with trigonometric equations. Establishing the fact that the equation is trigonometric is not difficult at all. Difficulties arise in determining the sequence of actions that would lead to the correct answer.
The appearance of an equation can sometimes be difficult to determine its type. And without knowing the type of equation, it is almost impossible to choose the desired one from several tens of trigonometric formulas.
To solve the trigonometric equation, one should try:
1. bring all the functions included in the equation to "equal angles";
2. to bring the equation to "the same functions";
3. factor 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 a trigonometric function in terms of known components.
Step 2. Find the argument of a function by the formulas:
cos x = a; x = ± arccos a + 2πn, n ЄZ.
sin x = a; x = (-1) n arcsin a + πn, n Є Z.
tg x = a; x = arctan a + πn, n Є Z.
ctg x = a; x = arcctg a + πn, n Є Z.
Step 3. Find 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 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 scheme
Step 1. Replace the given equation with a linear one, using the degree reduction formulas for this:
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 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 mind
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 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, 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 = -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 the equation solved by methods I, II, III, IV.
Step 2. Solve the resulting equation by 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 to solve trigonometric equations is very important, their development requires significant efforts, both on the part of the student and on the part of the teacher.
Many problems of stereometry, physics, etc. are connected with the solution of trigonometric equations. The process of solving such problems, as it were, contains many 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 the development of personality in general.
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Methods for solving trigonometric equations
Introduction 2
Methods for solving trigonometric equations 5
Algebraic 5
Solving Equations Using the Equality Condition for Trigonometric Functions of the Same Name 7
Factoring 8
Reduction to homogeneous equation 10
Auxiliary corner introduction 11
Convert work to sum 14
Universal Substitution 14
Conclusion 17
Introduction
Until the tenth grade, the order of actions of many exercises leading to the goal, as a rule, is unambiguously defined. For example, linear and quadratic equations and inequalities, fractional equations and equations reducible to quadratic, etc. Without examining in detail the principle of solving each of the above examples, let us note what is common that is necessary for their successful solution.
In most cases, it is necessary to establish what type of task the task belongs to, recall the sequence of actions leading to the goal, and perform these actions. Obviously, the success or failure of a student in mastering the methods of solving equations depends mainly on how well he is able to correctly determine the type of equation and remember the sequence of all stages of its solution. Of course, this assumes that the student has the skills to perform identical transformations and calculations.
A completely different situation occurs when a student encounters trigonometric equations. At the same time, it is not difficult to establish the fact that the equation is trigonometric. Difficulties arise when finding an order of actions that would lead to a positive result. And here the student faces two problems. It is difficult to determine the type from the appearance of the equation. And without knowing the type, it is almost impossible to choose the right formula from several dozen available.
To help students find the right path in the complex maze of trigonometric equations, they are first introduced to equations that, after introducing a new variable, are reduced to square ones. Then the homogeneous equations are solved and reduced to them. Everything ends, as a rule, with equations, for the solution of which it is necessary to factor the left side, then equating each of the factors to zero.
Realizing that the one and a half dozen equations analyzed in the lessons is clearly not enough to start the student on an independent voyage on the trigonometric "sea", the teacher adds a few more recommendations from himself.
To solve the trigonometric equation, one should try:
Reduce all functions included in the equation to "equal angles";
Reduce the equation to "identical functions";
Factor the left side of the equation, etc.
But, despite the knowledge of the basic types of trigonometric equations and several principles for finding their solution, many students still find themselves in a dead end before each equation that is slightly different from those that were solved before. It remains unclear what one should strive for, having this or that equation, why in one case it is necessary to apply the formulas of the double angle, in the other - half, and in the third - the formulas for addition, etc.
Definition 1. Trigonometric is an equation in which the unknown is contained under the sign of trigonometric functions.
Definition 2. They say that a trigonometric equation has the same angles if all trigonometric functions included in it have equal arguments. A trigonometric equation is said to have the same functions if it contains only one of the trigonometric functions.
Definition 3. The degree of a monomial containing trigonometric functions is the sum of the exponents of the powers of the trigonometric functions included in it.
Definition 4. An equation is called homogeneous if all monomials included in it have the same degree. This degree is called the order of the equation.
Definition 5. Trigonometric Equation Containing Only Functions sin and cos, is called homogeneous if all monomials with respect to trigonometric functions have the same degree, and the trigonometric functions themselves have equal angles and the number of monomials is 1 more than the order of the equation.
Methods for solving trigonometric equations.
Solving trigonometric equations consists of two stages: transforming the equation to obtain its simplest form and solving the resulting simplest trigonometric equation. There are seven basic methods for solving trigonometric equations.
I. Algebraic method. This method is well known from algebra. (Variable substitution and substitution method).
Solve equations.
1)
Let us introduce the notation x=2 sin3 t, we get
Solving this equation, we get: 
or 
those. can be written
When recording the received decision due to the presence of signs 
degree 
it makes no sense to write down.
Answer: 
We denote 
We get the quadratic equation 
... Its roots are numbers 
and 
... Therefore, this equation reduces to the simplest trigonometric equations 
and 
... Solving them, we find that 
or 
.
Answer: 
; 
.
We denote 
does not satisfy the condition 
Means 
Answer: 
Let's transform the left side of the equation:
Thus, this initial equation can be written as:
, i.e.
By designating 
, we get 
Having solved this quadratic equation, we have:
does not satisfy the condition 
We write down the solution to the original equation:
Answer: 
Substitution 
reduces this equation to a quadratic equation 
... Its roots are numbers 
and 
... Because 
, then the given equation has no roots.
Answer: there are no roots.
II... Solution of equations using the condition of equality of the same trigonometric functions.
a) 
, if
b) 
, if
v) 
, if 
Using these conditions, consider the solution of the following equations:
6) 
Using what was said in part a), we find that the equation has a solution if and only if 
.
Solving this equation, we find 
.
We have two groups of solutions: 
.
7) Solve the equation: 
.
Using condition b), we deduce that 
.
Solving these quadratic equations, we get:
.
8) Solve the equation 
.
From this equation, we deduce that. Solving this quadratic equation, we find that 
.
III... Factorization.
We consider this method by examples.
9) Solve the equation 
.
Solution. Move all the terms of the equation to the left:.
Transform and factorize the expression on the left side of the equation: 
.
.
.
1)
2) 
Because 
and 
do not take the value zero
at the same time, then we divide both parts
equations for 
, 
Answer: 
10) Solve the equation:
Solution.
or 
Answer: 
11) Solve the equation
Solution:
1)
2)
3)
, 
Answer: 
IV... Reduction to a homogeneous equation.
To solve a homogeneous equation you need:
Move all its members to the left side;
Move all common factors out of parentheses;
Set all factors and parentheses to zero;
The parentheses equated to zero give a homogeneous equation of lesser degree, which should be divided by 
(or 
) in the senior degree;
Solve the resulting algebraic equation for 
.
Let's consider some examples:
12) Solve the equation:
Solution.
Divide both sides of the equation by 
, 
Introducing the notation 
, named
roots of this equation: 
hence 1) 
2) 
Answer: 
13) Solve the equation:
Solution. Using the double angle formulas and the basic trigonometric identity, we reduce this equation to a half argument:
After reducing similar terms, we have:
Dividing the last homogeneous equation by 
, we get
I will designate 
, we get the quadratic equation 
whose roots are the numbers 
Thus 
Expression 
vanishes at 
, i.e. at 
, 
.
Our solution to the equation does not include these numbers.
Answer: 
, .
V... Introduction of an auxiliary angle.
Consider an equation of the form
Where a, b, c- coefficients, x- the unknown.
We divide both sides of this equation by 
Now the coefficients of the equation have the properties of sine and cosine, namely: the modulus of each of them does not exceed one, and the sum of their squares is 1.
Then we can denote them accordingly 
(here 
- auxiliary angle) and our equation takes the form:.
Then 
And his decision
Note that the introduced designations are mutually interchangeable.
14) Solve the equation: 
Solution. Here 
, so we divide both sides of the equation by 
Answer: 
15) Solve the equation
Solution. Because 
, then this equation is equivalent to the equation
Because 
, then there is an angle such that 
, 
(those. 
).
We have 
Because 
, then we finally get:
.
Note that an equation of the form has a solution if and only if 
16) Solve the equation:
To solve this equation, we group trigonometric functions with the same arguments
Divide both sides of the equation by two
We transform the sum of trigonometric functions into a product:
Answer: 
VI... Converting a work to a sum.
The corresponding formulas are used here.
17) Solve the equation:
Solution. Convert the left side to the sum:
Vii.Generic substitution.
, 
these formulas are true for everyone 
Substitution 
called universal.
18) Solve the equation: 
Solution: Replace and 
to their expression through 
and denote 
.
We get the rational equation 
which converts to square 
.
The roots of this equation are the numbers 
.
Therefore, the problem was reduced to solving two equations 
.
We find that 
.
View value 
does not satisfy the original equation, which is verified by checking - substitution of this value t into the original equation.
Answer: 
.
Comment. Equation 18 could be solved in a different way.
Divide both sides of this equation by 5 (i.e., by 
): 
.
Because 
, then there is such a number 
, what 
and 
... Therefore, the equation takes the form: 
or 
... From this we find that 
where 
.
19) Solve the equation 
.
Solution. Since the functions 
and 
have the greatest value equal to 1, then their sum is equal to 2, if 
and 
, simultaneously, that is 
.
Answer: 
.
When solving this equation, the boundedness of the functions and was used.
Conclusion.
Working on the topic "Solutions of trigonometric equations", it is useful for each teacher to follow these recommendations:
To systematize methods for solving trigonometric equations.
Choose for yourself the steps for performing the analysis of the equation and signs of the appropriateness of using one or another solution method.
Think over the ways of self-control of their activities for the implementation of the method.
Learn to compose "your" equations for each of the studied methods.
Appendix # 1
Solve homogeneous or homogeneous equations.
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