What is the difference between the simplest trigonometric equations. Trigonometric Equations
Examples:
\ (2 \ sin (x) = \ sqrt (3) \)
tg \ ((3x) = - \) \ (\ frac (1) (\ sqrt (3)) \)
\ (4 \ cos ^ 2x + 4 \ sinx-1 = 0 \)
\ (\ cos4x + 3 \ cos2x = 1 \)
How to solve trigonometric equations:
Any trigonometric equation should be reduced to one of the following types:
\ (\ sint = a \), \ (\ cost = a \), tg \ (t = a \), ctg \ (t = a \)
where \ (t \) is an expression with an x, \ (a \) is a number. Such trigonometric equations are called the simplest... They can be easily solved using () or special formulas:
Example ... Solve the trigonometric equation \ (\ sinx = - \) \ (\ frac (1) (2) \).
Solution:
Answer: \ (\ left [\ begin (gathered) x = - \ frac (π) (6) + 2πk, \\ x = - \ frac (5π) (6) + 2πn, \ end (gathered) \ right. \) \ (k, n∈Z \)
For what each symbol in the formula for the roots of trigonometric equations means, see.
Attention! The equations \ (\ sinx = a \) and \ (\ cosx = a \) have no solutions if \ (a ϵ (-∞; -1) ∪ (1; ∞) \). Because sine and cosine for any x is greater than or equal to \ (- 1 \) and less than or equal to \ (1 \):
\ (- 1≤ \ sin x≤1 \) \ (- 1≤ \ cosx≤1 \)
Example
... Solve the equation \ (\ cosx = -1,1 \).
Solution:
\(-1,1<-1\), а значение косинуса не может быть меньше \(-1\). Значит у уравнения нет решения.
Answer
: no solutions.
Example ... Solve the trigonometric equation tg \ (x = 1 \).
Solution:
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Let's solve the equation using the number circle. For this: |
Example
... Solve the trig equation \ (\ cos (3x + \ frac (π) (4)) = 0 \).
Solution:
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Let's use the number circle again. \ (3x + \) \ (\ frac (π) (4) \) \ (= ± \) \ (\ frac (π) (2) \) \ (+ 2πk \), \ (k∈Z \) \ (3x + \) \ (\ frac (π) (4) \) \ (= \) \ (\ frac (π) (2) \) \ (+ 2πk \) \ (3x + \) \ (\ frac ( π) (4) \) \ (= - \) \ (\ frac (π) (2) \) \ (+ 2πk \) 8) As usual, we will express \ (x \) in the equations. \ (3x = - \) \ (\ frac (π) (4) \) \ (+ \) \ (\ frac (π) (2) \) \ (+ 2πk \) \ (3x = - \) \ (\ frac (π) (4) \) \ (+ \) \ (\ frac (π) (2) \) \ (+ 2πk \) |
Reducing trigonometric equations to the simplest is a creative task, here you need to use and, and special methods for solving equations:
- Method (the most popular in the exam).
- Method.
- Method of auxiliary arguments.
Consider an example of solving the square-trigonometric equation
Example ... Solve the trig equation \ (2 \ cos ^ 2x-5 \ cosx + 2 = 0 \)Solution:
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\ (2 \ cos ^ 2x-5 \ cosx + 2 = 0 \) |
Let's make the replacement \ (t = \ cosx \). |
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Our equation has become typical. You can solve it with. |
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\ (D = 25-4 \ cdot 2 \ cdot 2 = 25-16 = 9 \) |
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\ (t_1 = \) \ (\ frac (5-3) (4) \) \ (= \) \ (\ frac (1) (2) \); \ (t_2 = \) \ (\ frac (5 + 3) (4) \) \ (= 2 \) |
We do the reverse replacement. |
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\ (\ cosx = \) \ (\ frac (1) (2) \); \ (\ cosx = 2 \) |
Solve the first equation using the number circle. |
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Let's write down all the numbers lying on at these points. |
An example of solving a trigonometric equation with the study of ODZ:
Example (exam) ... Solve the trigonometric equation \ (= 0 \)|
\ (\ frac (2 \ cos ^ 2x- \ sin (2x)) (ctg x) \)\(=0\) |
If there is a fraction and there is a cotangent, then you need to write it down. Let me remind you that the cotangent is actually a fraction: ctg \ (x = \) \ (\ frac (\ cosx) (\ sinx) \) Therefore, ODZ for ctg \ (x \): \ (\ sinx ≠ 0 \). |
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ODZ: ctg \ (x ≠ 0 \); \ (\ sinx ≠ 0 \)
\ (x ≠ ± \) \ (\ frac (π) (2) \) \ (+ 2πk \); \ (x ≠ πn \); \ (k, n∈Z \) |
Let's mark the “non-solutions” on the number circle. |
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\ (\ frac (2 \ cos ^ 2x- \ sin (2x)) (ctg x) \)\(=0\) |
Let's get rid of the denominator in the equation by multiplying it by ctg \ (x \). We can do this, since we wrote above that ctg \ (x ≠ 0 \). |
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\ (2 \ cos ^ 2x- \ sin (2x) = 0 \) |
Apply the double angle sine formula: \ (\ sin (2x) = 2 \ sinx \ cosx \). |
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\ (2 \ cos ^ 2x-2 \ sinx \ cosx = 0 \) |
If your hands are stretched out to divide by cosine - pull them back! You can divide by an expression with a variable if it is not exactly zero (for example, such as \ (x ^ 2 + 1.5 ^ x \)). Instead, put \ (\ cosx \) outside the parentheses. |
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\ (\ cosx (2 \ cosx-2 \ sinx) = 0 \) |
Let's "split" the equation into two. |
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\ (\ cosx = 0 \); \ (2 \ cosx-2 \ sinx = 0 \) |
Solve the first equation with a number circle. Divide the second equation by \ (2 \) and move \ (\ sinx \) to the right side. |
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\ (x = ± \) \ (\ frac (π) (2) \) \ (+ 2πk \), \ (k∈Z \). \ (\ cosx = \ sinx \) |
The roots that turned out are not included in the LDZ. Therefore, we will not write them down in response. |
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Use a circle again. |
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These roots are not excluded by the ODZ, so you can write them in response. |
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 X.
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 X , only put back in the negative direction. This is the corner -X. 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 X 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 X equal to the angle X ... 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 X 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 instance:
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.
Solving the simplest trigonometric equations.
The solution of trigonometric equations of any level of complexity ultimately comes down to solving the simplest trigonometric equations. And in this, the trigonometric circle turns out to be the best helper again.
Let's recall the definitions of cosine and sine.
The cosine of an angle is the abscissa (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation by a given angle.
The sine of an angle is the ordinate (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation by a given angle.
The positive direction of movement in the trigonometric circle is counterclockwise movement. A rotation of 0 degrees (or 0 radians) corresponds to a point with coordinates (1; 0)
We will use these definitions to solve the simplest trigonometric equations.
1. Let's solve the equation
This equation is satisfied by all such values of the angle of rotation, which correspond to the points of the circle, the ordinate of which is equal to.
Let's mark on the ordinate axis a point with an ordinate:
Let's draw a horizontal line parallel to the abscissa axis until it intersects with the circle. We get two points lying on a circle and having an ordinate. These points correspond to the angles of rotation by and radians:
If we, leaving the point corresponding to the angle of rotation by radians, go around a full circle, then we will come to the point corresponding to the angle of rotation by radians and having the same ordinate. That is, this angle of rotation also satisfies our equation. We can do as many "idle" revolutions as we want, returning to the same point, and all these values of the angles will satisfy our equation. The number of "idle" revolutions will be denoted by the letter (or). Since we can make these revolutions both in the positive and in the negative direction, (or) can take any integer values.
That is, the first series of solutions to the original equation has the form:
,, is the set of integers (1)
Similarly, the second series of solutions is:
, where , . (2)
As you may have guessed, this series of solutions is based on the point of the circle corresponding to the angle of rotation by.
These two series of solutions can be combined into one entry:
If we take in this record (that is, even), then we get the first series of solutions.
If we take in this record (that is, odd), then we get the second series of solutions.
2. Now let's solve the equation
Since is the abscissa of the point of the unit circle obtained by turning through an angle, mark the point with the abscissa on the axis:
Draw a vertical line parallel to the axis until it intersects with the circle. We get two points lying on a circle and having an abscissa. These points correspond to the angles of rotation by and radians. Recall that when moving clockwise, we get a negative rotation angle:
Let's write down two series of solutions:
,
,
(We get to the desired point, passing from the main full circle, that is.
Let's combine these two series into one entry:
3. Solve the equation
The tangent line passes through the point with coordinates (1,0) of the unit circle parallel to the OY axis
We mark a point on it with an ordinate equal to 1 (we are looking for the tangent of which angles is 1):
Let's connect this point with the origin of coordinates with a straight line and mark the points of intersection of the straight line with the unit circle. The intersection points of the straight line and the circle correspond to the angles of rotation on and:
Since the points corresponding to the angles of rotation that satisfy our equation lie at a distance of radians from each other, we can write the solution in this way:
4. Solve the equation
The cotangent line passes through the point with the coordinates of the unit circle parallel to the axis.
Let's mark on the line of cotangents a point with abscissa -1:
Let's connect this point with the origin of coordinates of a straight line and continue it to the intersection with the circle. This line will intersect the circle at the points corresponding to the angles of rotation by and radians:
Since these points are at a distance equal to each other, we can write the general solution of this equation as follows:
In the given examples, illustrating the solution of the simplest trigonometric equations, tabular values of trigonometric functions were used.
However, if there is not a tabular value on the right side of the equation, then we substitute the value in the general solution of the equation:
SPECIAL SOLUTIONS:
Let us mark on the circle the points whose ordinate is equal to 0:
Let us mark on the circle a single point, the ordinate of which is equal to 1:
Let's mark on the circle the only point, the ordinate of which is equal to -1:
Since it is customary to indicate the values that are closest to zero, we write the solution as follows:
Note on the circle the points whose abscissa is equal to 0:
5.
Let's mark on the circle the only point, the abscissa of which is equal to 1:
Let's mark on the circle the only point, the abscissa of which is equal to -1:
And a little more complex examples:
1.
The sine is one if the argument is
The argument of our sine is equal, so we get:
Divide both sides of the equality by 3:
Answer:
2.
Cosine is zero if the argument of the cosine is
The argument of our cosine is equal, so we get:
Let us express, for this we first move to the right with the opposite sign:
Let's simplify the right side:
Divide both parts by -2:
Note that the sign does not change in front of the term, since k can take any integer values.
Answer:
And finally, watch the video tutorial "Selecting roots in a trigonometric equation using a trigonometric circle"
This concludes the conversation about solving the simplest trigonometric equations. Next time we'll talk about how to solve.
Many math problems, especially those that occur before grade 10, the order of actions performed that will lead to the goal is clearly defined. These 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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