Trigonometric equations. The simplest trigonometric equations Solve the trigonometric equation sinx 1 2

The main methods of solving trigonometric equations are: reduction of equations to the simplest (using trigonometric formulas), introduction of new variables, factorization. Let's consider their application with examples. Pay attention to the registration of the solution of trigonometric equations.

Necessary condition successful solution trigonometric equations is the knowledge of trigonometric formulas (topic 13 of work 6).

Examples.

1. Equations Reducing to the Simplest.

1) Solve the equation

Decision:

Answer:

2) Find the roots of the equation

(sinx + cosx) 2 = 1 – sinxcosx belonging to the segment .

Decision:

Answer:

2. Equations Reducing to Quadratic Equations.

1) Solve the equation 2 sin 2 x - cosx -1 = 0.

Decision: Using the formula sin 2 x \u003d 1 - cos 2 x, we get

Answer:

2) Decide cos equation 2x = 1 + 4 cosx.

Decision: Using the formula cos 2x = 2 cos 2 x - 1, we get

Answer:

3) Decide tgx equation– 2ctgx + 1 = 0

Decision:

Answer:

3. Homogeneous equations

1) Solve the equation 2sinx - 3cosx = 0

Solution: Let cosx = 0, then 2sinx = 0 and sinx = 0 - a contradiction with the fact that sin 2 x + cos 2 x = 1. So cosx ≠ 0 and you can divide the equation by cosx. Get

Answer:

2) Solve the equation 1 + 7 cos 2 x = 3 sin 2x

Decision:

Using the formulas 1 = sin 2 x + cos 2 x and sin 2x = 2 sinxcosx, we get

sin2x + cos2x + 7cos2x = 6sinxcosx
sin2x - 6sinxcosx+ 8cos2x = 0

Let cosx = 0, then sin 2 x = 0 and sinx = 0 - a contradiction with the fact that sin 2 x + cos 2 x = 1.
So cosx ≠ 0 and we can divide the equation by cos 2 x . Get

tg 2x – 6 tgx + 8 = 0
Denote tgx = y
y 2 – 6 y + 8 = 0
y 1 = 4; y2=2
a) tanx = 4, x= arctg4 + 2 k, k
b) tgx = 2, x= arctg2 + 2 k, k .

Answer: arctg4 + 2 k, arctan2 + 2 k, k

4. Equations of the form a sinx + b cosx = with, with≠ 0.

1) Solve the equation.

Decision:

Answer:

5. Equations Solved by Factorization.

1) Solve the equation sin2x - sinx = 0.

The root of the equation f (X) = φ ( X) can only serve as the number 0. Let's check this:

cos 0 = 0 + 1 - the equality is true.

The number 0 is the only root of this equation.

Answer: 0.

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The simplest trigonometric equations are usually solved by formulas. Let me remind you that the following trigonometric equations are called the simplest:

sinx = a

cosx = a

tgx = a

ctgx = a

x is the angle to be found,
a is any number.

And here are the formulas with which you can immediately write down the solutions of these simplest equations.

For sinus:

For cosine:

x = ± arccos a + 2π n, n ∈ Z

For tangent:

x = arctg a + π n, n ∈ Z

For cotangent:

x = arcctg a + π n, n ∈ Z

Actually, this is theoretical part solutions of the simplest trigonometric equations. And, the whole!) Nothing at all. However, the number of errors on this topic just rolls over. Especially, with a slight deviation of the example from the template. Why?

Yes, because a lot of people write down these letters, without understanding their meaning at all! With apprehension he writes down, no matter how something happens ...) This needs to be dealt with. Trigonometry for people, or people for trigonometry, after all!?)

Let's figure it out?

One angle will be equal to arccos a, second: -arccos a.

And that's how it will always work. For any a.

If you don't believe me, hover your mouse over the picture, or touch the picture on the tablet.) I changed the number a to some negative. Anyway, we got one corner arccos a, second: -arccos a.

Therefore, the answer can always be written as two series of roots:

x 1 = arccos a + 2π n, n ∈ Z

x 2 = - arccos a + 2π n, n ∈ Z

We combine these two series into one:

x= ± arccos a + 2π n, n ∈ Z

And all things. We have obtained a general formula for solving the simplest trigonometric equation with cosine.

If you understand that this is not some kind of super-scientific wisdom, but just an abbreviated record of two series of answers, you and tasks "C" will be on the shoulder. With inequalities, with the selection of roots from a given interval ... There, the answer with plus / minus does not roll. And if you treat the answer businesslike, and break it into two separate answers, everything is decided.) Actually, for this we understand. What, how and where.

In the simplest trigonometric equation

sinx = a

also get two series of roots. Always. And these two series can also be recorded one line. Only this line will be smarter:

x = (-1) n arcsin a + π n, n ∈ Z

But the essence remains the same. Mathematicians simply constructed a formula to make one instead of two records of series of roots. And that's it!

Let's check the mathematicians? And that's not enough...)

In the previous lesson, the solution (without any formulas) of the trigonometric equation with a sine was analyzed in detail:

The answer turned out to be two series of roots:

x 1 = π /6 + 2π n, n ∈ Z

x 2 = 5π /6 + 2π n, n ∈ Z

If we solve the same equation using the formula, we get the answer:

x = (-1) n arcsin 0.5 + π n, n ∈ Z

Actually, this is a half-finished answer.) The student must know that arcsin 0.5 = π /6. The full answer would be:

x = (-1) n π /6+ πn, n ∈ Z

Here an interesting question arises. Reply via x 1; x 2 (this is the correct answer!) and through the lonely X (and this is the correct answer!) - the same thing, or not? Let's find out now.)

Substitute in response with x 1 values n =0; one; 2; etc., we consider, we get a series of roots:

x 1 \u003d π / 6; 13π/6; 25π/6 etc.

With the same substitution in response to x 2 , we get:

x 2 \u003d 5π / 6; 17π/6; 29π/6 etc.

And now we substitute the values n (0; 1; 2; 3; 4...) into the general formula for the lonely X . That is, we raise minus one to the zero power, then to the first, second, and so on. And, of course, we substitute 0 into the second term; one; 2 3; 4 etc. And we think. We get a series:

x = π/6; 5π/6; 13π/6; 17π/6; 25π/6 etc.

That's all you can see.) The general formula gives us exactly the same results which are the two answers separately. All at once, in order. Mathematicians did not deceive.)

Formulas for solving trigonometric equations with tangent and cotangent can also be checked. But let's not.) They are so unpretentious.

I painted all this substitution and verification on purpose. Here it is important to understand one simple thing: there are formulas for solving elementary trigonometric equations, just a summary of the answers. For this brevity, I had to insert plus/minus into the cosine solution and (-1) n into the sine solution.

These inserts do not interfere in any way in tasks where you just need to write down the answer to an elementary equation. But if you need to solve an inequality, or then you need to do something with the answer: select roots on an interval, check for ODZ, etc., these inserts can easily unsettle a person.

And what to do? Yes, either paint the answer in two series, or solve the equation / inequality in a trigonometric circle. Then these inserts disappear and life becomes easier.)

You can sum up.

To solve the simplest trigonometric equations, there are ready-made answer formulas. Four pieces. They are good for instantly writing the solution to an equation. For example, you need to solve the equations:

sinx = 0.3

Easily: x = (-1) n arcsin 0.3 + π n, n ∈ Z

cosx = 0.2

No problem: x = ± arccos 0.2 + 2π n, n ∈ Z

tgx = 1.2

Easily: x = arctg 1,2 + πn, n ∈ Z

ctgx = 3.7

One left: x= arcctg3,7 + πn, n ∈ Z

cos x = 1.8

If you, shining with knowledge, instantly write the answer:

x= ± arccos 1.8 + 2π n, n ∈ Z

then you already shine, this ... that ... from a puddle.) The correct answer is: there are no solutions. Don't understand why? Read what an arccosine is. In addition, if on the right side of the original equation there are tabular values \u200b\u200bof sine, cosine, tangent, cotangent, - 1; 0; √3; 1/2; √3/2 etc. - the answer through the arches will be unfinished. Arches must be converted to radians.

And if you already come across an inequality, like

then the answer is:

x πn, n ∈ Z

there is a rare nonsense, yes ...) Here it is necessary to trigonometric circle decide. What we will do in the corresponding topic.

For those who heroically read up to these lines. I just can't help but appreciate your titanic efforts. you a bonus.)

Bonus:

When writing formulas in an anxious combat situation, even hardened nerds often get confused where pn, And where 2πn. Here's a simple trick for you. In all formulas pn. Except for the only formula with arc cosine. It stands there 2πn. Two pien. Keyword - two. In the same single formula are two sign at the beginning. Plus and minus. Here and there - two.

So if you wrote two sign in front of the arc cosine, it is easier to remember what will happen at the end two pien. And vice versa happens. Skip the man sign ± , get to the end, write correctly two pien, yes, and catch it. Ahead of something two sign! The person will return to the beginning, but he will correct the mistake! Like this.)

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

Once I witnessed a conversation between two applicants:

– When do you need to add 2πn, and when - πn? I can't remember!

- And I have the same problem.

I wanted to say to them: “It is not necessary to memorize, but to understand!”

This article is addressed primarily to high school students and, I hope, will help them with "understanding" to solve the simplest trigonometric equations:

Number circle

Along with the concept of a number line, there is also the concept number circle. As we know, in a rectangular coordinate system, a circle with a center at the point (0; 0) and a radius of 1 is called a unit circle. Imagine a number line with a thin thread and wind it around this circle: the reference point (point 0), attach it to the “right” point unit circle, we wrap the positive semi-axis counterclockwise, and the negative one - in the direction (Fig. 1). Such a unit circle is called a number circle.

Number circle properties

Every real number is at one point on the number circle.
There are infinitely many real numbers on each point of the number circle. Since the length of the unit circle is 2π, the difference between any two numbers at one point on the circle is equal to one of the numbers ±2π; ±4π; ±6π; …

Let's conclude: knowing one of the numbers of point A, we can find all the numbers of point A.

Let's draw the AC diameter (Fig. 2). Since x_0 is one of the numbers of the point A, then the numbers x_0±π ; x_0±3π; x_0±5π; … and only they will be the numbers of the point C. Let's choose one of these numbers, say, x_0+π, and use it to write down all the numbers of the point C: x_C=x_0+π+2πk ,k∈Z. Note that the numbers at points A and C can be combined into one formula: x_(A ; C)=x_0+πk ,k∈Z (for k = 0; ±2; ±4; ... we get the numbers of point A, and for k = ±1, ±3, ±5, … are the numbers of the point C).

Let's conclude: knowing one of the numbers on one of the points A or C of the diameter AC, we can find all the numbers on these points.

Two opposite numbers are located on points of the circle that are symmetrical about the abscissa axis.

Let's draw a vertical chord AB (Fig. 2). Since points A and B are symmetrical about the Ox axis, the number -x_0 is located at point B and, therefore, all numbers of point B are given by the formula: x_B=-x_0+2πk ,k∈Z. We write the numbers at points A and B with one formula: x_(A ; B)=±x_0+2πk ,k∈Z. Let's conclude: knowing one of the numbers at one of the points A or B of the vertical chord AB, we can find all the numbers at these points. Consider the horizontal chord AD and find the numbers of the point D (Fig. 2). Since BD is the diameter and the number -x_0 belongs to point B, then -x_0 + π is one of the numbers of point D and, therefore, all numbers of this point are given by the formula x_D=-x_0+π+2πk ,k∈Z. Numbers at points A and D can be written using one formula: x_(A ; D)=(-1)^k∙x_0+πk ,k∈Z . (for k= 0; ±2; ±4; ... we get the numbers of point A, and for k = ±1; ±3; ±5; ... - the numbers of point D).

Let's conclude: knowing one of the numbers at one of the points A or D of the horizontal chord AD, we can find all the numbers at these points.

Sixteen main points of the number circle

In practice, the solution of most of the simplest trigonometric equations is associated with sixteen points of the circle (Fig. 3). What are these dots? Red, blue and green dots divide the circle into 12 equal parts. Since the length of the semicircle is π, the length of the arc A1A2 is π/2, the length of the arc A1B1 is π/6, and the length of the arc A1C1 is π/3.

Now we can specify one number on the points:

π/3 on С1 and

The vertices of the orange square are the midpoints of the arcs of each quarter, so the length of the arc A1D1 is equal to π/4, and hence π/4 is one of the numbers of the point D1. Using the properties of the number circle, we can write down all the numbers at all the marked points of our circle using formulas. The figure also shows the coordinates of these points (we omit the description of their acquisition).

Having learned the above, we now have sufficient preparation for solving special cases (for nine values of the number a) the simplest equations.

Solve Equations

1)sinx=1⁄(2).

– What is required of us?

– Find all those numbers x whose sine is 1/2.

Recall the definition of sine: sinx - the ordinate of the point of the number circle, on which the number x is located. On the circle we have two points, the ordinate of which is equal to 1/2. These are the ends of the horizontal chord B1B2. This means that the requirement “solve the equation sinx=1⁄2” is equivalent to the requirement “find all numbers at point B1 and all numbers at point B2”.

2)sinx=-√3⁄2 .

We need to find all the numbers at the points C4 and C3.

3) sinx=1. On the circle we have only one point with ordinate 1 - point A2 and, therefore, we need to find only all the numbers of this point.

Answer: x=π/2+2πk , k∈Z .

4)sinx=-1 .

Only point A_4 has ordinate -1. All numbers of this point will be the horses of the equation.

Answer: x=-π/2+2πk , k∈Z .

5) sinx=0 .

On the circle we have two points with ordinate 0 - points A1 and A3. You can specify the numbers on each of the points separately, but given that these points are diametrically opposed, it is better to combine them into one formula: x=πk ,k∈Z .

Answer: x=πk ,k∈Z .

6)cosx=√2⁄2 .

Recall the definition of cosine: cosx - abscissa of the point of the numerical circle on which the number x is located. On the circle we have two points with the abscissa √2⁄2 - the ends of the horizontal chord D1D4. We need to find all the numbers at these points. We write them down by combining them into one formula.

Answer: x=±π/4+2πk , k∈Z .

7) cosx=-1⁄2 .

We need to find the numbers at the points C_2 and C_3 .

Answer: x=±2π/3+2πk , k∈Z .

10) cosx=0 .

Only points A2 and A4 have abscissa 0, which means that all numbers at each of these points will be solutions to the equation.
.

The solutions of the equation of the system are the numbers at the points B_3 and B_4. Inequality cosx<0 удовлетворяют только числа b_3
Answer: x=-5π/6+2πk , k∈Z .

Note that for any admissible value of x, the second factor is positive and, therefore, the equation is equivalent to the system

The solutions of the system equation are the number of points D_2 and D_3 . The numbers of the point D_2 do not satisfy the inequality sinx≤0.5, but the numbers of the point D_3 do.