Trigonometric equations and inequalities calculator. Simple and complex trigonometric inequalities

On practical lesson We will repeat the main types of tasks from the topic “Trigonometry”, additionally analyze problems of increased complexity and consider examples of solving various trigonometric inequalities and their systems.

This lesson will help you prepare for one of the types of tasks B5, B7, C1 and C3.

Let's start by reviewing the main types of tasks that we covered in the topic "Trigonometry" and solve several non-standard problems.

Task No. 1. Convert angles to radians and degrees: a) ; b) .

a) Let’s use the formula for converting degrees to radians

Let's substitute it specified value.

b) Apply the formula for converting radians to degrees

Let's perform the substitution .

Answer. A) ; b) .

Task No. 2. Calculate: a) ; b) .

a) Since the angle goes far beyond the table, we will reduce it by subtracting the sine period. Because The angle is indicated in radians, then we will consider the period as .

b) B in this case the situation is similar. Since the angle is indicated in degrees, we will consider the period of the tangent as .

The resulting angle, although smaller than the period, is larger, which means that it no longer refers to the main, but to the extended part of the table. In order not to once again train your memory by memorizing the extended table of trigofunction values, let’s subtract the tangent period again:

We took advantage of the oddness of the tangent function.

Answer. a) 1; b) .

Task No. 3. Calculate , If .

Let us reduce the entire expression to tangents by dividing the numerator and denominator of the fraction by . At the same time, we can not be afraid that, because in this case, the tangent value would not exist.

Task No. 4. Simplify the expression.

The specified expressions are converted using reduction formulas. They are just unusually written using degrees. The first expression generally represents a number. Let's simplify all the trigofunctions one by one:

Because , then the function changes to a cofunction, i.e. to the cotangent, and the angle falls into the second quarter, in which the original tangent has a negative sign.

For the same reasons as in the previous expression, the function changes to a cofunction, i.e. to the cotangent, and the angle falls into the first quarter, in which the original tangent has a positive sign.

Let's substitute everything into a simplified expression:

Problem #5. Simplify the expression.

Let's write down the tangent double angle using the appropriate formula and simplify the expression:

The last identity is one of the universal replacement formulas for the cosine.

Problem #6. Calculate.

The main thing is not to make the standard mistake and not give the answer that the expression is equal to . You cannot use the basic property of the arctangent as long as there is a factor in the form of two next to it. To get rid of it, we will write the expression according to the formula for the tangent of a double angle, while treating , as an ordinary argument.

Now we can apply the basic property of the arctangent; remember that there are no restrictions on its numerical result.

Problem No. 7. Solve the equation.

When deciding fractional equation, which is equal to zero, it is always indicated that the numerator is equal to zero, but the denominator is not, because You cannot divide by zero.

The first equation is a special case of the simplest equation that can be solved using a trigonometric circle. Remember this solution yourself. The second inequality is solved as the simplest equation using the general formula for the roots of the tangent, but only with the sign not equal to.

As we see, one family of roots excludes another family of exactly the same type of roots that do not satisfy the equation. Those. there are no roots.

Answer. There are no roots.

Problem No. 8. Solve the equation.

Let’s immediately note that we can take out the common factor and let’s do it:

The equation has been reduced to one of the standard forms, when the product of several factors equals zero. We already know that in this case, either one of them is equal to zero, or the other, or the third. Let's write this in the form of a set of equations:

The first two equations are special cases of the simplest ones; we have already encountered similar equations many times, so we will immediately indicate their solutions. We reduce the third equation to one function using the double angle sine formula.

Let's solve the last equation separately:

This equation has no roots, because the sine value cannot go beyond .

Thus, the solution is only the first two families of roots; they can be combined into one, which is easy to show on the trigonometric circle:

This is a family of all halves, i.e.

Let's move on to solving trigonometric inequalities. First, let's look at the approach to solving the example without using formulas general solutions, but using the trigonometric circle.

Problem No. 9. Solve inequality.

Let us draw an auxiliary line on the trigonometric circle corresponding to a sine value equal to , and show the range of angles that satisfy the inequality.

It is very important to understand exactly how to indicate the resulting interval of angles, i.e. what is its beginning and what is its end. The beginning of the interval will be the angle corresponding to the point that we will enter at the very beginning of the interval if we move counterclockwise. In our case, this is the point that is on the left, because moving counterclockwise and passing right point, on the contrary, we are leaving the required range of angles. The right point will therefore correspond to the end of the gap.

Now we need to understand the angles of the beginning and end of our interval of solutions to the inequality. Common mistake- this is to immediately indicate that the right point corresponds to the angle, the left one and give the answer. This is not true! Please note that we have just indicated the interval corresponding to the upper part of the circle, although we are interested in the lower part, in other words, we have mixed up the beginning and end of the solution interval we need.

In order for the interval to start from the corner of the right point and end with the corner of the left point, it is necessary that the first specified angle be less than the second. To do this, we will have to measure the angle of the right point in the negative direction of reference, i.e. clockwise and it will be equal to . Then, starting to move from it in a positive clockwise direction, we will get to the right point after the left point and get the angle value for it. Now the beginning of the interval of angles is less than the end, and we can write the interval of solutions without taking into account the period:

Considering that such intervals will be repeated an infinite number of times after any integer number of rotations, we obtain a general solution taking into account the sine period:

We put parentheses because the inequality is strict, and we pick out the points on the circle that correspond to the ends of the interval.

Compare the answer you receive with the formula for the general solution that we gave in the lecture.

Answer. .

This method is good for understanding where the formulas for general solutions of the simplest trigon inequalities come from. In addition, it is useful for those who are too lazy to learn all these cumbersome formulas. However, the method itself is also not easy; choose which approach to the solution is most convenient for you.

To solve trigonometric inequalities, you can also use graphs of functions on which an auxiliary line is constructed, similar to the method shown using a unit circle. If you are interested, try to figure out this approach to the solution yourself. In the future we will use general formulas for solving simple trigonometric inequalities.

Problem No. 10. Solve inequality.

Let us use the formula for the general solution, taking into account the fact that the inequality is not strict:

In our case we get:

Answer.

Problem No. 11. Solve inequality.

Let us use the general solution formula for the corresponding strictly inequality:

Answer. .

Problem No. 12. Solve inequalities: a) ; b) .

In these inequalities, there is no need to rush to use formulas for general solutions or the trigonometric circle; it is enough to simply remember the range of values ​​of sine and cosine.

a) Since , then the inequality does not make sense. Therefore, there are no solutions.

b) Because similarly, the sine of any argument always satisfies the inequality specified in the condition. Therefore, the inequality is satisfied by all real values ​​of the argument.

Answer. a) there are no solutions; b) .

Problem 13. Solve inequality .

1. If the argument is complex (different from X), then replace it with t.

2. We build in one coordinate plane tOy function graphs y=cost And y=a.

3. We find such two adjacent points of intersection of graphs, between which is located above the straight line y=a. We find the abscissas of these points.

4. Write a double inequality for the argument t, taking into account the cosine period ( t will be between the found abscissas).

5. Let's do it reverse replacement(return to the original argument) and express the value X from double inequality, write the answer in the form of a numerical interval.

Example 1.

Next, according to the algorithm, we determine those values ​​of the argument t, at which the sinusoid is located higher direct. Let's write these values ​​as a double inequality, taking into account the periodicity of the cosine function, and then return to the original argument X.

Example 2.

Selecting a range of values t, in which the sinusoid is above the straight line.

We write the values ​​in the form of double inequality t, satisfying the condition. Do not forget that the smallest period of the function y=cost equals 2π. Returning to the variable X, gradually simplifying all parts of the double inequality.

We write the answer in the form of a closed numerical interval, since the inequality was not strict.

Example 3.

We will be interested in the range of values t, at which the points of the sinusoid will lie above the straight line.

Values t write it in the form of a double inequality, rewrite the same values ​​for 2x and express X. Let's write the answer in the form of a numerical interval.

And again formula cost>a.

If cost>a, (-1≤A≤1), then - arccos a + 2πn< t < arccos a + 2πn, nєZ.

Apply formulas to solve trigonometric inequalities and you will save time on exam testing.

And now formula , which you should use in the UNT or Unified State Examination when deciding trigonometric inequality kind cost

If cost , (-1≤A≤1), then arccos a + 2πn< t < 2π — arccos a + 2πn, nєZ.

Apply this formula to solve the inequalities discussed in this article, and you will get the answer much faster and without any graphs!

Taking into account the periodicity of the sine function, we write a double inequality for the values ​​of the argument t, satisfying the last inequality. Let's return to the original variable. Let us transform the resulting double inequality and express the variable X. Let's write the answer in the form of an interval.

Let's solve the second inequality:

When solving the second inequality, we had to transform the left side of this inequality using the double argument sine formula to obtain an inequality of the form: sint≥a. Next we followed the algorithm.

We solve the third inequality:

Dear graduates and applicants! Keep in mind that methods for solving trigonometric inequalities, such as the graphical method given above and, probably known to you, the method of solving using a unit trigonometric circle (trigonometric circle) are applicable only in the first stages of studying the section of trigonometry “Solving trigonometric equations and inequalities.” I think you will remember that you first solved the simplest trigonometric equations using graphs or a circle. However, now you wouldn't think of solving trigonometric equations this way. How do you solve them? That's right, according to the formulas. So trigonometric inequalities should be solved using formulas, especially during testing, when every minute is precious. So, solve the three inequalities of this lesson using the appropriate formula.

If sint>a, where -1≤ a≤1, then arcsin a + 2πn< t < π — arcsin a + 2πn, nєZ.

Learn formulas!

And finally: did you know that mathematics is definitions, rules and FORMULAS?!

Of course you do! And the most curious, having studied this article and watched the video, exclaimed: “How long and difficult! Is there a formula that allows you to solve such inequalities without any graphs or circles?” Yes, of course there is!

TO SOLVING INEQUALITIES OF THE FORM: sin (-1≤A≤1) the formula is valid:

— π — arcsin a + 2πn< t < arcsin a + 2πn, nєZ.

Apply it to the examples discussed and you will get the answer much faster!

Conclusion: LEARN FORMULAS, FRIENDS!

Page 1 of 1 1

Inequalities are relations of the form a › b, where a and b are expressions containing at least one variable. Inequalities can be strict - ‹, › and non-strict - ≥, ≤.

Trigonometric inequalities are expressions of the form: F(x) › a, F(x) ‹ a, F(x) ≤ a, F(x) ≥ a, in which F(x) is represented by one or more trigonometric functions.

An example of the simplest trigonometric inequality is: sin x ‹ 1/2. It is customary to solve such problems graphically; two methods have been developed for this.

Method 1 - Solving inequalities by graphing a function

To find an interval that satisfies the conditions of inequality sin x ‹ 1/2, you must perform the following steps:

1. On the coordinate axis, construct a sinusoid y = sin x.
2. On the same axis, draw a graph of the numerical argument of the inequality, i.e., a straight line passing through the point ½ of the ordinate OY.
3. Mark the intersection points of the two graphs.
4. Shade the segment that is the solution to the example.

When strict signs are present in an expression, the intersection points are not solutions. Since the smallest positive period of a sinusoid is 2π, we write the answer as follows:

If the signs of the expression are not strict, then the solution interval must be enclosed in square brackets - . The answer to the problem can also be written as the following inequality:

Method 2 - Solving trigonometric inequalities using the unit circle

Similar problems can be easily solved using a trigonometric circle. The algorithm for finding answers is very simple:

1. First you need to draw a unit circle.
2. Then you need to note the value of the arc function of the argument of the right side of the inequality on the arc of the circle.
3. It is necessary to draw a straight line passing through the value of the arc function parallel to the abscissa axis (OX).
4. After that, all that remains is to select the arc of a circle, which is the set of solutions to the trigonometric inequality.
5. Write down the answer in the required form.

Let us analyze the stages of the solution using the example of the inequality sin x › 1/2. Points α and β are marked on the circle - values

The points of the arc located above α and β are the interval for solving the given inequality.

If you need to solve an example for cos, then the answer arc will be located symmetrically to the OX axis, not OY. You can consider the difference between the solution intervals for sin and cos in the diagrams below in the text.

Graphical solutions for tangent and cotangent inequalities will differ from both sine and cosine. This is due to the properties of functions.

Arctangent and arccotangent are tangents to a trigonometric circle, and the minimum positive period for both functions is π. To quickly and correctly use the second method, you need to remember on which axis the values ​​of sin, cos, tg and ctg are plotted.

The tangent tangent runs parallel to the OY axis. If we plot the value of arctan a on the unit circle, then the second required point will be located in the diagonal quarter. Angles

They are break points for the function, since the graph tends to them, but never reaches them.

In the case of cotangent, the tangent runs parallel to the OX axis, and the function is interrupted at points π and 2π.

Complex trigonometric inequalities

If the argument of the inequality function is represented not just by a variable, but by an entire expression containing an unknown, then we are talking about a complex inequality. The process and procedure for solving it are somewhat different from the methods described above. Suppose we need to find a solution to the following inequality:

The graphical solution involves constructing an ordinary sinusoid y = sin x using arbitrarily selected values ​​of x. Let's calculate a table with coordinates for the control points of the graph:

The result should be a beautiful curve.

To make finding a solution easier, let’s replace the complex function argument