Solution of 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 again turns out to be the best helper.

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 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 rotation by a given angle.

The positive direction of movement trigonometric circle counterclockwise movement is considered. A rotation of 0 degrees (or 0 radians) corresponds to a point with coordinates (1; 0)

We use these definitions to solve the simplest trigonometric equations.

1. 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 a point with ordinate on the y-axis:

Draw a horizontal line parallel to the x-axis until it intersects with the circle. We will get two points lying on a circle and having an ordinate. These points correspond to rotation angles of and radians:

If we, having left the point corresponding to the angle of rotation per radian, go around a full circle, then we will come to a point corresponding to the angle of rotation per radian and having the same ordinate. That is, this angle of rotation also satisfies our equation. We can make as many "idle" turns as we like, returning to the same point, and all these angle values ​​will satisfy our equation. The number of "idle" revolutions is denoted by the letter (or). Since we can make these revolutions in both positive and negative directions, (or ) can take on any integer values.

That is, the first series of solutions to the original equation has the form:

, , - set of integers (1)

Similarly, the second series of solutions has the form:

, Where , . (2)

As you 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 entry (that is, even), then we will get the first series of solutions.

If we take in this entry (that is, odd), then we will 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 the angle , we mark on the axis a point with the abscissa :

Draw a vertical line parallel to the axis until it intersects with the circle. We will get two points lying on a circle and having an abscissa. These points correspond to rotation angles of and radians. Recall that when moving clockwise, we get a negative angle of rotation:

We write down two series of solutions:

,

,

(We get to the right point by passing from the main full circle, that is.

Let's combine these two series into one post:

3. Solve the equation

The line of tangents passes through the point with coordinates (1,0) of the unit circle parallel to the OY axis

Mark a point on it with an ordinate equal to 1 (we are looking for the tangent of which angles is 1):

Connect this point to the origin with a straight line and mark the points of intersection of the line with the unit circle. The points of intersection of the line and the circle correspond to the rotation angles on and :

Since the points corresponding to the rotation angles that satisfy our equation lie radians apart, we can write the solution as follows:

4. Solve the equation

The line of cotangents passes through the point with the coordinates of the unit circle parallel to the axis.

We mark a point with the abscissa -1 on the line of cotangents:

Connect this point to the origin of the straight line and continue it until it intersects with the circle. This line will intersect the circle at points corresponding to rotation angles of and radians:

Since these points are separated from each other by a distance equal to , then 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 ​​were used trigonometric functions.

However, if there is a non-table value on the right side of the equation, then we substitute the value in the general solution of the equation:

SPECIAL SOLUTIONS:

Mark points on the circle whose ordinate is 0:

Mark a single point on the circle, the ordinate of which is equal to 1:

Mark a single point on the circle, the ordinate of which is equal to -1:

Since it is customary to indicate the values ​​​​closest to zero, we write the solution as follows:

Mark the points on the circle, the abscissa of which is 0:

5.
Let's mark a single point on the circle, the abscissa of which is equal to 1:

Mark a single point on the circle, the abscissa of which is equal to -1:

And some more complex examples:

1.

The sine is one if the argument is

The argument of our sine is , so we get:

Divide both sides of the equation by 3:

Answer:

2.

The cosine is zero if the cosine argument is

The argument of our cosine is , so we get:

We express , for this we first move to the right with the opposite sign:

Simplify the right side:

Divide both parts by -2:

Note that the sign before the term does not change, since k can take any integer values.

Answer:

And in conclusion, watch the video tutorial "Selection of 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.

Trigonometric equations are not the easiest topic. Painfully they are diverse.) For example, these:

sin2x + cos3x = ctg5x

sin(5x+π /4) = ctg(2x-π /3)

sinx + cos2x + tg3x = ctg4x

Etc...

But these (and all other) trigonometric monsters have two common and obligatory features. First - you won't believe it - there are trigonometric functions in the equations.) Second: all expressions with x are within these same functions. And only there! If x appears somewhere outside, For example, sin2x + 3x = 3, this will be a mixed type equation. Such equations require an individual approach. Here we will not consider them.

We will not solve evil equations in this lesson either.) Here we will deal with the simplest trigonometric equations. Why? Yes, because the decision any trigonometric equations consists of two stages. At the first stage, the evil equation is reduced to a simple one by various transformations. On the second - this simplest equation is solved. No other way.

So, if you have problems in 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 stands for any number. Any.

By the way, inside the function there may be not a pure x, but some kind of expression, such as:

cos(3x+π /3) = 1/2

etc. This complicates life, but does not affect the method of solving the trigonometric equation.

How to solve trigonometric equations?

Trigonometric equations can be solved in two ways. The first way: using logic and a trigonometric circle. We will explore this path here. The second way - using memory and formulas - will be considered 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!

We solve equations using a trigonometric circle.

We include elementary logic and the ability to use a trigonometric circle. Can't you!? However... It will be 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 textbooks...)

Ah, you know!? And even mastered "Practical work with a trigonometric circle"!? Accept congratulations. This topic will be close and understandable to you.) What is especially pleasing is that the trigonometric circle does not care which equation you solve. Sine, cosine, tangent, cotangent - everything is the same for him. The solution principle is the same.

So we take any elementary trigonometric equation. At least this:

cosx = 0.5

I need to find X. If to speak human language, need to find the angle (x) whose cosine is 0.5.

How did we use the circle before? We drew a corner on it. In degrees or radians. And immediately seen trigonometric functions of this angle. Now let's do the opposite. Draw a cosine equal to 0.5 on the circle and immediately we'll see corner. It remains only to write down the answer.) Yes, yes!

We draw a circle and mark the cosine equal to 0.5. On the cosine axis, of course. Like this:

Now let's draw the angle that this cosine gives us. Hover your mouse over the picture (or touch the picture on a tablet), and see this same corner X.

Which angle has a cosine of 0.5?

x \u003d π / 3

cos 60°= cos( π /3) = 0,5

Some people will grunt skeptically, yes... They say, was it worth it to fence the circle, when everything is clear anyway... You can, of course, grunt...) But the fact is that this is an erroneous answer. Or rather, inadequate. Connoisseurs of the circle understand that there are still a whole bunch of angles that also give a cosine equal to 0.5.

If you turn the movable side OA for a 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 a solution to our equation, because

There are an infinite number of such full rotations... And all these new angles will be solutions to our trigonometric equation. And they all need to be written down somehow. All. Otherwise, the decision is not considered, yes ...)

Mathematics can do this simply and elegantly. In one short answer, write down infinite set solutions. Here's what it looks like for our equation:

x = π /3 + 2π n, n ∈ Z

I will decipher. Still write meaningfully nicer than stupidly drawing some mysterious letters, right?)

π /3 is the same angle that we saw on the circle and determined according to the table of cosines.

2π is one full turn in radians.

n - this is the number of complete, i.e. whole revolutions. It is clear that n can be 0, ±1, ±2, ±3.... and so on. As indicated by the short entry:

n ∈ Z

n belongs ( ∈ ) to the set of integers ( Z ). By the way, instead of the letter n letters can be used k, m, t etc.

This notation means that you can take any integer n . At least -3, at least 0, at least +55. What do you want. If you plug that number into your answer, you get a specific angle, which is sure to be the solution to our harsh equation.)

Or, in other words, x \u003d π / 3 is the only root of an infinite set. To get all the other roots, it is enough to add any number of full turns to π / 3 ( n ) in radians. Those. 2πn radian.

All? No. I specifically stretch the pleasure. To remember better.) We received only a 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 other angles that also give a cosine equal to 0.5!

Let's return to our picture, according to which we wrote down the answer. Here she is:

Move the mouse over the image and see another corner that also gives a cosine of 0.5. What do you think it equals? The triangles are the same... Yes! He equal to the angle X , only plotted in the negative direction. This is the corner -X. But we have already calculated x. π /3 or 60°. Therefore, we can safely write:

x 2 \u003d - π / 3

And, of course, we add all the angles that are obtained through full turns:

x 2 = - π /3 + 2π n, n ∈ Z

That's all now.) In a trigonometric circle, we saw(who understands, of course)) All angles that give a cosine equal to 0.5. And they wrote down these angles in a short mathematical form. The answer is two infinite 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 for solving trigonometric equations with the help of a circle is understandable. We mark on the circle the cosine (sine, tangent, cotangent) from given equation, draw the corners 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 so obvious. Well, as I said, logic is required here.)

For example, let's analyze 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. We draw a circle, mark (on the sine axis, of course!) 0.5. We draw at once all the angles corresponding to this sine. We get this picture:

Let's deal with the angle first. X in the first quarter. We recall the table of sines and determine the value of this angle. The matter is simple:

x \u003d π / 6

We recall full turns and, with a clear conscience, write down the first series of answers:

x 1 = π /6 + 2π n, n ∈ Z

Half the job is done. Now we need to define second corner... This is trickier 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 counted from the angle π in the negative direction. That's why it's red.) And for the answer, we need an angle measured correctly from the positive semiaxis OX, 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 of interest to us (drawn in green) will be equal to:

π - x

x we know it π /6 . So the second angle will be:

π - π /6 = 5π /6

Again, we recall the addition of full revolutions and write down the second series of answers:

x 2 = 5π /6 + 2π n, n ∈ Z

That's all. A 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. Unless, of course, you know how to draw the tangent and cotangent on a trigonometric circle.

In the examples above, I used the tabular value of sine and cosine: 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 the following trigonometric equation:

This cosine value in summary tables No. We coolly ignore this terrible fact. We draw a circle, mark 2/3 on the cosine axis and draw the corresponding angles. We get this picture.

We understand, for starters, with an angle in the first quarter. To know what x is equal to, they would immediately write down the answer! We don't know... Failure!? Calm! Mathematics does not leave its own in trouble! She invented arc cosines for this case. Do not know? In vain. Find out. It's a lot easier than you think. According to this link, there is not a single tricky spell about "inverse trigonometric functions" ... It's superfluous in this topic.

If you're in the know, just say to yourself, "X is an angle whose cosine is 2/3." And immediately, purely by definition of the arccosine, we can write:

We remember about additional revolutions 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 written 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 all things! This is the correct answer. Even easier than with tabular values. You don’t need to remember anything.) By the way, the most attentive will notice that this picture with the solution through the arc cosine is essentially no different from the picture for the equation cosx = 0.5.

Exactly! The general principle on that and the general! I specifically drew two almost identical pictures. The circle shows us the angle X by its cosine. It is a tabular cosine, or not - the circle does not know. What kind of angle is this, π / 3, or what kind of arc cosine is up to us to decide.

With a sine the same song. For example:

Again we draw a circle, mark the sine equal to 1/3, draw the corners. It turns out this picture:

And again the picture is almost the same as for the equation sinx = 0.5. Again we start from the corner in the first quarter. What is x equal to 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

Let's take a look at the second angle. In the example with a table value of 0.5, it was equal to:

π - x

So here it will be exactly the same! Only x is different, arcsin 1/3. So what!? You can safely write the second pack of roots:

x 2 = π - arcsin 1/3 + 2π n, n ∈ Z

This is a completely correct answer. Although it does not look very familiar. But it's 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 on a given interval, in trigonometric inequalities - they are generally solved almost always in a circle. In short, in any tasks that are a little more complicated than standard ones.

Putting knowledge into practice?

Solve trigonometric equations:

At first it is simpler, directly on this lesson.

Now it's more difficult.

Hint: here you have to think about the circle. Personally.)

And now 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 there is one ... And how to write down one instead of two series of answers. Yes, so that not a single root from an infinite number is lost!)

Well, quite simple):

sinx = 0,3

cosx = π

tgx = 1,2

ctgx = 3,7

Hint: here you need to know what is the arcsine, arccosine? What is arc tangent, arc tangent? The simplest definitions. But you don’t need to remember any tabular values!)

The answers are, of course, in disarray):

x 1= arcsin0,3 + 2πn, n ∈ Z
x 2= π - arcsin0.3 + 2

Not everything works out? Happens. Read the lesson again. Only thoughtfully(there is such an obsolete word...) And follow the links. The main links are about the circle. Without it in trigonometry - how to cross the road blindfolded. 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. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

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Lesson and presentation on the topic: "Solution of the simplest trigonometric equations"

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What will we study:
1. What are trigonometric equations?

3. Two main methods for solving trigonometric equations.
4. Homogeneous trigonometric equations.
5. Examples.

What are trigonometric equations?

Guys, we have already studied the arcsine, arccosine, arctangent and arccotangent. Now let's look at trigonometric equations in general.

Trigonometric equations - equations in which the variable is contained under the sign of the trigonometric function.

We repeat the form of solving the simplest trigonometric equations:

1) If |а|≤ 1, then the equation cos(x) = a has a solution:

X= ± arccos(a) + 2πk

2) If |а|≤ 1, then the equation sin(x) = a has a solution:

3) If |a| > 1, then the equation sin(x) = a and cos(x) = a have no solutions 4) The equation tg(x)=a has a solution: x=arctg(a)+ πk

5) The equation ctg(x)=a has a solution: x=arcctg(a)+ πk

For all formulas, k is an integer

The simplest trigonometric equations have the form: Т(kx+m)=a, T- any trigonometric function.

Example.

Solve equations: a) sin(3x)= √3/2

Solution:

A) Let's denote 3x=t, then we will rewrite our equation in the form:

The solution to this equation will be: t=((-1)^n)arcsin(√3/2)+ πn.

From the table of values ​​we get: t=((-1)^n)×π/3+ πn.

Let's go back to our variable: 3x =((-1)^n)×π/3+ πn,

Then x= ((-1)^n)×π/9+ πn/3

Answer: x= ((-1)^n)×π/9+ πn/3, where n is an integer. (-1)^n - minus one to the power of n.

More examples of trigonometric equations.

Solve the equations: a) cos(x/5)=1 b)tg(3x- π/3)= √3

Solution:

A) This time we will go directly to the calculation of the roots of the equation right away:

X/5= ± arccos(1) + 2πk. Then x/5= πk => x=5πk

Answer: x=5πk, where k is an integer.

B) We write in the form: 3x- π/3=arctg(√3)+ πk. We know that: arctg(√3)= π/3

3x- π/3= π/3+ πk => 3x=2π/3 + πk => x=2π/9 + πk/3

Answer: x=2π/9 + πk/3, where k is an integer.

Solve equations: cos(4x)= √2/2. And find all the roots on the segment .

Solution:

We will decide in general view our equation: 4x= ± arccos(√2/2) + 2πk

4x= ± π/4 + 2πk;

X= ± π/16+ πk/2;

Now let's see what roots fall on our segment. For k For k=0, x= π/16, we are in the given segment .
With k=1, x= π/16+ π/2=9π/16, they hit again.
For k=2, x= π/16+ π=17π/16, but here we didn’t hit, which means we won’t hit for large k either.

Answer: x= π/16, x= 9π/16

Two main solution methods.

We have considered the simplest trigonometric equations, but there are more complex ones. To solve them, the method of introducing a new variable and the factorization method are used. Let's look at examples.

Let's solve the equation:

Solution:
To solve our equation, we use the method of introducing a new variable, denoted: t=tg(x).

As a result of the replacement, we get: t 2 + 2t -1 = 0

Let's find the roots quadratic equation: t=-1 and t=1/3

Then tg(x)=-1 and tg(x)=1/3, we got the simplest trigonometric equation, let's find its roots.

X=arctg(-1) +πk= -π/4+πk; x=arctg(1/3) + πk.

Answer: x= -π/4+πk; x=arctg(1/3) + πk.

An example of solving an equation

Solve equations: 2sin 2 (x) + 3 cos(x) = 0

Solution:

Let's use the identity: sin 2 (x) + cos 2 (x)=1

Our equation becomes: 2-2cos 2 (x) + 3 cos (x) = 0

2 cos 2 (x) - 3 cos(x) -2 = 0

Let's introduce the replacement t=cos(x): 2t 2 -3t - 2 = 0

The solution to our quadratic equation are the roots: t=2 and t=-1/2

Then cos(x)=2 and cos(x)=-1/2.

Because cosine cannot take values ​​greater than one, then cos(x)=2 has no roots.

For cos(x)=-1/2: x= ± arccos(-1/2) + 2πk; x= ±2π/3 + 2πk

Answer: x= ±2π/3 + 2πk

Homogeneous trigonometric equations.

Definition: An equation of the form a sin(x)+b cos(x) is called homogeneous trigonometric equations of the first degree.

Equations of the form

homogeneous trigonometric equations of the second degree.

To solve a homogeneous trigonometric equation of the first degree, we divide it by cos(x): It is impossible to divide by cosine if it is equal to zero, let's make sure that this is not so:
Let cos(x)=0, then asin(x)+0=0 => sin(x)=0, but sine and cosine are not equal to zero at the same time, we got a contradiction, so we can safely divide by zero.

Solve the equation:
Example: cos 2 (x) + sin(x) cos(x) = 0

Solution:

Take out the common factor: cos(x)(c0s(x) + sin (x)) = 0

Then we need to solve two equations:

cos(x)=0 and cos(x)+sin(x)=0

Cos(x)=0 for x= π/2 + πk;

Consider the equation cos(x)+sin(x)=0 Divide our equation by cos(x):

1+tg(x)=0 => tg(x)=-1 => x=arctg(-1) +πk= -π/4+πk

Answer: x= π/2 + πk and x= -π/4+πk

How to solve homogeneous trigonometric equations of the second degree?
Guys, stick to these rules always!

1. See what the coefficient a is equal to, if a \u003d 0 then our equation will take the form cos (x) (bsin (x) + ccos (x)), an example of the solution of which is on the previous slide

2. If a≠0, then you need to divide both parts of the equation by the squared cosine, we get:

We make the change of variable t=tg(x) we get the equation:

Solve Example #:3

Solve the equation:
Solution:

Divide both sides of the equation by cosine square:

We make a change of variable t=tg(x): t 2 + 2 t - 3 = 0

Find the roots of the quadratic equation: t=-3 and t=1

Then: tg(x)=-3 => x=arctg(-3) + πk=-arctg(3) + πk

Tg(x)=1 => x= π/4+ πk

Answer: x=-arctg(3) + πk and x= π/4+ πk

Solve Example #:4

Solve the equation:

Solution:
Let's transform our expression:

We can solve such equations: x= - π/4 + 2πk and x=5π/4 + 2πk

Answer: x= - π/4 + 2πk and x=5π/4 + 2πk

Solve Example #:5

Solve the equation:

Solution:
Let's transform our expression:

We introduce the replacement tg(2x)=t:2 2 - 5t + 2 = 0

The solution to our quadratic equation will be the roots: t=-2 and t=1/2

Then we get: tg(2x)=-2 and tg(2x)=1/2
2x=-arctg(2)+ πk => x=-arctg(2)/2 + πk/2

2x= arctg(1/2) + πk => x=arctg(1/2)/2+ πk/2

Answer: x=-arctg(2)/2 + πk/2 and x=arctg(1/2)/2+ πk/2

Tasks for independent solution.

1) Solve the equation

A) sin(7x)= 1/2 b) cos(3x)= √3/2 c) cos(-x) = -1 d) tg(4x) = √3 e) ctg(0.5x) = -1.7

2) Solve equations: sin(3x)= √3/2. And find all the roots on the segment [π/2; π].

3) Solve the equation: ctg 2 (x) + 2ctg(x) + 1 =0

4) Solve the equation: 3 sin 2 (x) + √3sin (x) cos(x) = 0

5) Solve the equation: 3sin 2 (3x) + 10 sin(3x)cos(3x) + 3 cos 2 (3x) =0

6) Solve the equation: cos 2 (2x) -1 - cos(x) =√3/2 -sin 2 (2x)