Exercise.
Find the value of x at .

Solution.
To find the value of the function argument, at which it is equal to some value, means to determine for which arguments the value of the sine will be exactly the same as indicated in the condition.
In this case, we need to find out at what values ​​the value of the sine will be equal to 1/2. This can be done in several ways.
For example, use , by which to determine at what values ​​of x the sine function will be equal to 1/2.
Another way is to use . Let me remind you that the values ​​of the sines lie on the Oy axis.
The most common way is to use , especially when it comes to such standard values ​​for this function as 1/2.
In all cases, one should not forget about one of the most important properties of the sine - its period.
Let's find the value 1/2 for the sine in the table and see what arguments correspond to it. The arguments we are interested in are Pi / 6 and 5Pi / 6.
Write down all the roots that satisfy the given equation. To do this, we write down the unknown argument x of interest to us and one of the values ​​of the argument obtained from the table, that is, Pi / 6. Let's write down for it, taking into account the sine period, all the values ​​of the argument:

Let's take the second value and follow the same steps as in the previous case:

The complete solution to the original equation will be:
And
q can take the value of any integer.

On a trigonometric circle, in addition to angles in degrees, we observe.

More about radians:

A radian is defined as the angular value of an arc whose length is equal to its radius. Accordingly, since the circumference is , then it is obvious that the radian fits in the circle, that is

1 rad ≈ 57.295779513° ≈ 57°17′44.806″ ≈ 206265″.

Everyone knows that a radian is

So, for example, , a . That's how we Learn how to convert radians to angles.

Now vice versa let's convert degrees to radians.

Let's say we need to convert to radians. Will help us. We proceed as follows:

Since, radian, then fill in the table:

We train to find the values ​​​​of sine and cosine in a circle

Let's clarify the following.

Well, it’s good if we are asked to calculate, say, - usually there is no confusion here - everyone starts looking first on the circle.

And if they are asked to calculate, for example, ... Many, suddenly, begin to not understand where to look for this zero ... Often they look for it at the origin. Why?

1) Let's agree once and for all! What comes after or is argument=angle, and our corners are on the circle, don't look for them on the x axis!(It's just that individual points fall on both the circle and the axis ...) And the values ​​\u200b\u200bof the sines and cosines themselves - we are looking for on the axes!

2) And more! If we depart from the starting point counterclock-wise(the main direction of bypassing the trigonometric circle), then we set aside the positive values ​​of the angles, the angles increase as we move in that direction.

If we depart from the starting point clockwise, then we set aside the negative values ​​of the angles.

Example 1

Find value .

Solution:

We find on the circle. We project the point onto the sine axis (that is, we draw a perpendicular from the point to the sine axis (oy)).

We arrive at 0. Hence, .

Example 2

Find value .

Solution:

We find on the circle (we pass counterclockwise and more). We project a point onto the sine axis (and it already lies on the sinus axis).

We fall into -1 along the sine axis.

Note that behind the point "hidden" are points such as (we could go to the point marked as , clockwise, which means a minus sign appears), and infinitely many others.

One can make the following analogy:

Imagine a trigonometric circle as a stadium treadmill.

After all, you can end up at the “Flag” point, I start counterclockwise, running, say, 300 m. Or running, say, 100 m clockwise (we consider the length of the track is 400 m).

And you can also end up at the “Flag” point (after “start”) by running, say, 700 m, 1100 m, 1500 m, etc. counterclockwise. You can reach the Flag Point by running 500m or 900m, etc. clockwise from start.

Mentally expand the stadium's treadmill into a number line. Imagine where on this line there will be, for example, the values ​​300, 700, 1100, 1500, etc. We will see points on the number line, equidistant from each other. Let's turn back around. The dots “stick together” into one.

So it is with the trigonometric circle. Behind each point there are infinitely many others.

Let's say angles , , , etc. shown as a single dot. And the values ​​of the sine, cosine in them, of course, are the same. (Did you notice that we added/subtracted or? This is the period for the sine and cosine function.)

Example 3

Find value .

Solution:

Let's convert to degrees for simplicity.

(later, when you get used to the trigonometric circle, you will not need to convert radians to degrees):

We will move clockwise from the point Let's go half a circle () and more

We understand that the value of the sine coincides with the value of the sine and is equal to

Note that if we took, for example, or, etc., then we would get the same sine value.

Example 4

Find value .

Solution:

However, we will not convert radians to degrees, as in the previous example.

That is, we need to go counterclockwise half a circle and another quarter of a half circle and project the resulting point onto the cosine axis (horizontal axis).

Example 5

Find value .

Solution:

How to plot on a trigonometric circle?

If we pass or, yes, at least, we will still end up at the point that we designated as “start”. Therefore, you can immediately go to a point on the circle

Example 6

Find value .

Solution:

We will end up at a point (will lead us anyway to point zero). We project the point of the circle onto the cosine axis (see the trigonometric circle), we get into. That is .

Trigonometric circle - in your hands

You already understood that the main thing is to remember the values ​​\u200b\u200bof the trigonometric functions of the first quarter. In the remaining quarters, everything is similar, you just need to follow the signs. And I hope you will not forget the “chain-ladder” of the values ​​​​of trigonometric functions.

How to find tangent and cotangent values main angles.

After that, having become acquainted with the basic values ​​\u200b\u200bof the tangent and cotangent, you can pass

On an empty circle template. Train!

The sine values ​​are in the range [-1; 1], i.e. -1 ≤ sin α ≤ 1. Therefore, if |a| > 1, then the equation sin x = a has no roots. For example, the equation sin x = 2 has no roots.

Let's turn to some tasks.

Solve the equation sin x = 1/2.

Solution.

Note that sin x is the ordinate of the point of the unit circle, which is obtained as a result of the rotation of the point Р (1; 0) by the angle x around the origin.

An ordinate equal to ½ is present at two points of the circle M 1 and M 2.

Since 1/2 \u003d sin π / 6, then the point M 1 is obtained from the point P (1; 0) by turning through the angle x 1 \u003d π / 6, as well as through the angles x \u003d π / 6 + 2πk, where k \u003d +/-1, +/-2, …

The point M 2 is obtained from the point P (1; 0) as a result of turning through the angle x 2 = 5π/6, as well as through the angles x = 5π/6 + 2πk, where k = +/-1, +/-2, ... , i.e. at angles x = π – π/6 + 2πk, where k = +/-1, +/-2, ….

So, all the roots of the equation sin x = 1/2 can be found by the formulas x = π/6 + 2πk, x = π - π/6 + 2πk, where k € Z.

These formulas can be combined into one: x \u003d (-1) n π / 6 + πn, where n € Z (1).

Indeed, if n is an even number, i.e. n = 2k, then from formula (1) we obtain х = π/6 + 2πk, and if n is an odd number, i.e. n = 2k + 1, then from formula (1) we obtain х = π – π/6 + 2πk.

Answer. x \u003d (-1) n π / 6 + πn, where n € Z.

Solve the equation sin x = -1/2.

Solution.

The ordinate -1/2 have two points of the unit circle M 1 and M 2, where x 1 = -π/6, x 2 = -5π/6. Therefore, all the roots of the equation sin x = -1/2 can be found by the formulas x = -π/6 + 2πk, x = -5π/6 + 2πk, k ∈ Z.

We can combine these formulas into one: x \u003d (-1) n (-π / 6) + πn, n € Z (2).

Indeed, if n = 2k, then by formula (2) we obtain x = -π/6 + 2πk, and if n = 2k – 1, then by formula (2) we find x = -5π/6 + 2πk.

Answer. x \u003d (-1) n (-π / 6) + πn, n € Z.

Thus, each of the equations sin x = 1/2 and sin x = -1/2 has an infinite number of roots.

On the segment -π/2 ≤ x ≤ π/2, each of these equations has only one root:
x 1 \u003d π / 6 - the root of the equation sin x \u003d 1/2 and x 1 \u003d -π / 6 - the root of the equation sin x \u003d -1/2.

The number π/6 is called the arcsine of the number 1/2 and is written: arcsin 1/2 = π/6; the number -π/6 is called the arcsine of the number -1/2 and they write: arcsin (-1/2) = -π/6.

In general, the equation sin x \u003d a, where -1 ≤ a ≤ 1, on the segment -π / 2 ≤ x ≤ π / 2 has only one root. If a ≥ 0, then the root is enclosed in the interval; if a< 0, то в промежутке [-π/2; 0). Этот корень называют арксинусом числа а и обозначают arcsin а.

Thus, the arcsine of the number a € [–1; 1] such a number is called a € [–π/2; π/2], whose sine is a.

arcsin a = α if sin α = a and -π/2 ≤ x ≤ π/2 (3).

For example, arcsin √2/2 = π/4, since sin π/4 = √2/2 and – π/2 ≤ π/4 ≤ π/2;
arcsin (-√3/2) = -π/3, since sin (-π/3) = -√3/2 and – π/2 ≤ – π/3 ≤ π/2.

Similarly to how it was done when solving problems 1 and 2, it can be shown that the roots of the equation sin x = a, where |a| ≤ 1 are expressed by the formula

x \u003d (-1) n arcsin a + πn, n € Z (4).

We can also prove that for any a € [-1; 1] the formula arcsin (-a) = -arcsin a is valid.

From formula (4) it follows that the roots of the equation
sin x \u003d a for a \u003d 0, a \u003d 1, a \u003d -1 can be found using simpler formulas:

sin x \u003d 0 x \u003d πn, n € Z (5)

sin x \u003d 1 x \u003d π / 2 + 2πn, n € Z (6)

sin x \u003d -1 x \u003d -π / 2 + 2πn, n € Z (7)

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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 along the trigonometric circle is considered to be movement counterclockwise. 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 ​​of trigonometric functions were used.

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.