Trigonometric equations and inequalities definition. Methods for solving trigonometric inequalities

1.5 Trigonometric inequalities and methods for solving them

1.5.1 Solving simple trigonometric inequalities

Most authors of modern mathematics textbooks suggest starting to consider this topic by solving the simplest trigonometric inequalities. The principle of solving the simplest trigonometric inequalities is based on the knowledge and skills of determining on a trigonometric circle the values ​​of not only the main trigonometric angles, but also other values.

Meanwhile, the solution to inequalities of the form , , , can be carried out as follows: first we find some interval () on which this inequality is satisfied, and then write down the final answer by adding to the ends of the found interval a number that is a multiple of the period of the sine or cosine: ( ). In this case, the value is easy to find, because or . The search for meaning is based on students’ intuition, their ability to notice the equality of arcs or segments, taking advantage of the symmetry of individual parts of the sine or cosine graph. And this is sometimes beyond the capabilities of quite a large number of students. In order to overcome the noted difficulties, textbooks in recent years have used different approaches to solving simple trigonometric inequalities, but this has not resulted in any improvement in learning outcomes.

For a number of years, we have been quite successfully using formulas for the roots of the corresponding equations to find solutions to trigonometric inequalities.

We study this topic in the following way:

1. We build graphs and y = a, assuming that .

Then we write down the equation and its solution. Giving n 0; 1; 2, we find the three roots of the compiled equation: . The values ​​are the abscissa of three consecutive points of intersection of the graphs and y = a. It is obvious that the inequality always holds on the interval (), and the inequality always holds on the interval ().

By adding to the ends of these intervals a number that is a multiple of the period of the sine, in the first case we obtain a solution to the inequality in the form: ; and in the second case, a solution to the inequality in the form:

Only in contrast to the sine from the formula, which is a solution to the equation, for n = 0 we obtain two roots, and the third root for n = 1 in the form . And again, they are three consecutive abscissas of the points of intersection of the graphs and . In the interval () the inequality holds, in the interval () the inequality

Now it is not difficult to write down the solutions to the inequalities and . In the first case we get: ;

and in the second: .

Summarize. To solve the inequality or, you need to create the corresponding equation and solve it. From the resulting formula, find the roots of and , and write the answer to the inequality in the form: .

When solving inequalities , from the formula for the roots of the corresponding equation we find the roots and , and write the answer to the inequality in the form: .

This technique allows you to teach all students how to solve trigonometric inequalities, because This technique relies entirely on skills that students have a strong command of. These are the skills to solve simple problems and find the value of a variable using a formula. In addition, careful resolution under the guidance of a teacher becomes completely unnecessary. large quantity exercises in order to demonstrate all sorts of reasoning techniques depending on the sign of inequality, the value of the modulus of the number a and its sign. And the process of solving inequality itself becomes brief and, which is very important, uniform.

Another advantage of this method is that it allows you to easily solve inequalities even when the right side is not a table value of sine or cosine.

Let's demonstrate this with a specific example. Suppose we need to solve an inequality. Let's create the corresponding equation and solve it:

Let's find the values ​​of and .

When n = 1

When n = 2

We write down the final answer to this inequality:

In the considered example of solving the simplest trigonometric inequalities, there can be only one drawback - the presence of a certain amount of formalism. But if everything is assessed only from these positions, then it will be possible to accuse the formulas of the roots of the quadratic equation, and all formulas for solving trigonometric equations, and much more, of formalism.

Although the proposed method occupies a worthy place in the formation of skills in solving trigonometric inequalities, the importance and features of other methods for solving trigonometric inequalities cannot be underestimated. These include the interval method.

Let's consider its essence.

Set edited by A.G. Mordkovich, although you shouldn’t ignore the rest of the textbooks either. § 3. Methodology for teaching the topic “Trigonometric functions” in the course of algebra and beginnings of analysis In the study of trigonometric functions at school, two main stages can be distinguished: ü Initial acquaintance with trigonometric functions...

In carrying out the research, the following tasks were solved: 1) The current textbooks of algebra and the beginnings of mathematical analysis were analyzed to identify the methods presented in them for solving irrational equations and inequalities. The analysis allows us to draw the following conclusions: ·in secondary school, insufficient attention is paid to methods for solving various irrational equations, mainly...

METHODS FOR SOLVING TRIGONOMETRIC INEQUALITIES

Relevance. Historically, trigonometric equations and inequalities have been given a special place in the school curriculum. We can say that trigonometry is one of the most important sections of the school course and the entire mathematical science in general.

Trigonometric equations and inequalities occupy one of the central places in the secondary school mathematics course, both in terms of the content of educational material and the methods of educational and cognitive activity that can and should be formed during their study and applied to solving a large number of problems of a theoretical and applied nature .

Solving trigonometric equations and inequalities creates the prerequisites for systematizing students’ knowledge related to all educational material in trigonometry (for example, properties of trigonometric functions, methods of transforming trigonometric expressions, etc.) and makes it possible to establish effective connections with the studied material in algebra (equations, equivalence of equations, inequalities, identical transformations of algebraic expressions, etc.).

In other words, consideration of techniques for solving trigonometric equations and inequalities involves a kind of transfer of these skills to new content.

The significance of the theory and its numerous applications are proof of the relevance of the chosen topic. This in turn allows you to determine the goals, objectives and subject of research of the course work.

Purpose of the study: generalize the available types of trigonometric inequalities, basic and special methods for solving them, select a set of problems for solving trigonometric inequalities by schoolchildren.

Research objectives:

1. Based on an analysis of the available literature on the research topic, systematize the material.

2. Provide a set of tasks necessary to consolidate the topic “Trigonometric inequalities.”

Object of study are trigonometric inequalities in the school mathematics course.

Subject of study: types of trigonometric inequalities and methods for solving them.

Theoretical significance is to systematize the material.

Practical significance: application of theoretical knowledge in solving problems; analysis of the main common methods for solving trigonometric inequalities.

Research methods : analysis of scientific literature, synthesis and generalization of acquired knowledge, analysis of problem solving, search for optimal methods for solving inequalities.

§1. Types of trigonometric inequalities and basic methods for solving them

1.1. The simplest trigonometric inequalities

Two trigonometric expressions connected by the sign or > are called trigonometric inequalities.

Solving a trigonometric inequality means finding the set of values ​​of the unknowns included in the inequality for which the inequality is satisfied.

The main part of trigonometric inequalities is solved by reducing them to the simplest solution:

This may be a method of factorization, change of variable (
,
etc.), where the usual inequality is first solved, and then an inequality of the form
etc., or other methods.

The simplest inequalities can be solved in two ways: using the unit circle or graphically.

Letf(x  – one of the basic trigonometric functions. To solve the inequality
it is enough to find its solution on one period, i.e. on any segment whose length is equal to the period of the functionf  x  . Then the solution to the original inequality will be all foundx , as well as those values ​​that differ from those found by any integer number of periods of the function. In this case, it is convenient to use the graphical method.

Let us give an example of an algorithm for solving inequalities
(
) And
.

Algorithm for solving inequality
(
).

1. Formulate the definition of the sine of a numberx on the unit circle.

3. On the ordinate axis, mark the point with the coordinatea .

4. Draw a line parallel to the OX axis through this point and mark its intersection points with the circle.

5. Select an arc of a circle, all points of which have an ordinate less thana .

6. Indicate the direction of the round (counterclockwise) and write down the answer by adding the period of the function to the ends of the interval2πn ,
.

Algorithm for solving inequality
.

1. Formulate the definition of the tangent of a numberx on the unit circle.

2. Draw a unit circle.

3. Draw a line of tangents and mark a point with an ordinate on ita .

4. Connect this point with the origin and mark the point of intersection of the resulting segment with the unit circle.

5. Select an arc of a circle, all points of which have an ordinate on the tangent line less thana .

6. Indicate the direction of the traversal and write the answer taking into account the domain of definition of the function, adding a periodπn ,
(the number on the left of the entry is always less than the number on the right).

Graphic interpretation of solutions to the simplest equations and formulas for solving inequalities in general form are indicated in the appendix (Appendices 1 and 2).

Example 1. Solve the inequality
.

Draw a straight line on the unit circle
, which intersects the circle at points A and B.

All meaningsy on the interval NM is greater , all points of the AMB arc satisfy this inequality. At all rotation angles, large , but smaller ,
will take on values ​​greater (but not more than one).

Fig.1

Thus, the solution to the inequality will be all values ​​in the interval
, i.e.
. In order to obtain all solutions to this inequality, it is enough to add to the ends of this interval
, Where
, i.e.
,
. Note that the values
And
are the roots of the equation
,

those.
;
.

Answer:
,
.

1.2. Graphical method

In practice, the graphical method for solving trigonometric inequalities often turns out to be useful. Let us consider the essence of the method using the example of inequality
:

1. If the argument is complex (different fromX ), then replace it witht .

2. We build in one coordinate planetOy function graphs
And
.

3. We find suchtwo adjacent points of intersection of graphs, between whichsine wavelocatedhigher straight
. We find the abscissas of these points.

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

5. Make a reverse substitution (return to the original argument) and express the valueX from the double inequality, we write the answer in the form of a numerical interval.

Example 2. Solve inequality: .

When solving inequalities using the graphical method, it is necessary to construct graphs of functions as accurately as possible. Let's transform the inequality to the form:

Let's construct graphs of functions in one coordinate system
And
(Fig. 2).

Fig.2

The graphs of functions intersect at the pointA with coordinates
;
. In between
graph points
below the graph points
. And when
the function values ​​are the same. That's why
at
.

Answer:
.

1.3. Algebraic method

Quite often, the original trigonometric inequality can be reduced to an algebraic (rational or irrational) inequality through a well-chosen substitution. This method involves transforming an inequality, introducing a substitution or replacing a variable.

Let's look at specific examples of the application of this method.

Example 3. Reduction to the simplest form
.

(Fig. 3)

Fig.3

,
.

Answer:
,

Example 4. Solve inequality:

ODZ:
,
.

Using formulas:
,

Let's write the inequality in the form:
.

Or, believing
after simple transformations we get

,

,

.

Solving the last inequality using the interval method, we obtain:

Fig.4

, respectively
. Then from Fig. 4 follows
, Where
.

Fig.5

Answer:
,
.

1.4. Interval method

General scheme for solving trigonometric inequalities using the interval method:

Factor using trigonometric formulas.

Find the discontinuity points and zeros of the function and place them on the circle.

Take any pointTO (but not found earlier) and find out the sign of the product. If the product is positive, then place a point outside the unit circle on the ray corresponding to the angle. Otherwise, place the point inside the circle.

If a point occurs an even number of times, we call it a point of even multiplicity; if an odd number of times, we call it a point of odd multiplicity. Draw arcs as follows: start from a pointTO , if the next point is of odd multiplicity, then the arc intersects the circle at this point, but if the point is of even multiplicity, then it does not intersect.

Arcs behind the circle are positive intervals; inside the circle there are negative spaces.

Example 5. Solve inequality

,
.

Points of the first series:
.

Points of the second series:
.

Each point occurs an odd number of times, that is, all points are of odd multiplicity.

Let us find out the sign of the product at
: . Let's mark all the points on the unit circle (Fig. 6):

Rice. 6

Answer:
,
;
,
;
,
.

Example 6 . Solve the inequality.

Solution:

Let's find the zeros of the expression .

Receiveaem :

,
;

,
;

,
;

,
;

On the unit circle series valuesX 1 represented by dots
. SeriesX 2 gives points
. A seriesX 3 we get two points
. Finally, the seriesX 4 will represent points
. Let's plot all these points on the unit circle, indicating its multiplicity in parentheses next to each of them.

Let now the number will be equal. Let's make an estimate based on the sign:

So, full stopA should be selected on the ray forming the angle with beamOh, outside the unit circle. (Note that the auxiliary beamABOUT A It is not at all necessary to depict it in a drawing. DotA is chosen approximately.)

Now from the pointA draw a wavy continuous line sequentially to all marked points. And at points
our line goes from one area to another: if it was outside the unit circle, then it goes inside it. Approaching the point , the line returns to the inner region, since the multiplicity of this point is even. Similarly at the point (with even multiplicity) the line has to be turned to the outer region. So, we drew a certain picture shown in Fig. 7. It helps to highlight the desired areas on the unit circle. They are marked with a “+” sign.

Fig.7

Final answer:

Note. If a wavy line, after going around all the points marked on the unit circle, cannot be returned to the pointA , without crossing the circle in an “illegal” place, this means that an error was made in the solution, namely, an odd number of roots were missed.

Answer: .

§2. A set of problems for solving trigonometric inequalities

In the process of developing the ability of schoolchildren to solve trigonometric inequalities, 3 stages can also be distinguished.

1. preparatory,

2. developing the ability to solve simple trigonometric inequalities;

3. introduction of trigonometric inequalities of other types.

The purpose of the preparatory stage is that it is necessary to develop in schoolchildren the ability to use a trigonometric circle or graph to solve inequalities, namely:

Ability to solve simple inequalities of the form
,
,
,
, using the properties of the sine and cosine functions;

Ability to construct double inequalities for arcs of the number circle or for arcs of graphs of functions;

Ability to perform various transformations of trigonometric expressions.

It is recommended to implement this stage in the process of systematizing schoolchildren’s knowledge about the properties of trigonometric functions. The main means can be tasks offered to students and performed either under the guidance of a teacher or independently, as well as skills developed in solving trigonometric equations.

Here are examples of such tasks:

1 . Mark a point on the unit circle , If

.

2. In which quarter of the coordinate plane is the point located? , If equals:

3. Mark the points on the trigonometric circle , If:

4. Convert the expression to trigonometric functionsIquarters.

A)
, b)
, V)

5. Arc MR is given.M – middleI-th quarter,R – middleIIth quarter. Limit the value of a variablet for: (make a double inequality) a) arc MR; b) RM arcs.

6. Write down the double inequality for the selected sections of the graph:

Rice. 1

7. Solve inequalities
,
,
,
.

8. Convert Expression .

At the second stage of learning to solve trigonometric inequalities, we can offer the following recommendations related to the methodology for organizing student activities. In this case, it is necessary to focus on the students’ existing skills in working with a trigonometric circle or graph, formed while solving the simplest trigonometric equations.

Firstly, one can motivate the expediency of obtaining a general method for solving the simplest trigonometric inequalities by turning, for example, to an inequality of the form
. Using the knowledge and skills acquired at the preparatory stage, students will bring the proposed inequality to the form
, but may find it difficult to find a set of solutions to the resulting inequality, because It is impossible to solve it only using the properties of the sine function. This difficulty can be avoided by turning to the appropriate illustration (solving the equation graphically or using a unit circle).

Secondly, the teacher should draw students’ attention to different ways of completing the task, give an appropriate example of solving the inequality both graphically and using a trigonometric circle.

Let us consider the following solutions to the inequality
.

1. Solving the inequality using the unit circle.

In the first lesson on solving trigonometric inequalities, we will offer students a detailed solution algorithm, which in a step-by-step presentation reflects all the basic skills necessary to solve the inequality.

Step 1.Let's draw a unit circle and mark a point on the ordinate axis and draw a straight line through it parallel to the x-axis. This line will intersect the unit circle at two points. Each of these points represents numbers whose sine is equal to .

Step 2.This straight line divided the circle into two arcs. Let us select the one that depicts numbers that have a sine greater than . Naturally, this arc is located above the drawn straight line.

Rice. 2

Step 3.Select one of the ends of the marked arc. Let's write down one of the numbers that is represented by this point of the unit circle .

Step 4.In order to select the number corresponding to the second end of the selected arc, we “walk” along this arc from the named end to the other. At the same time, recall that when moving counterclockwise, the numbers we will go through increase (when moving in the opposite direction, the numbers would decrease). Let's write down the number that is depicted on the unit circle by the second end of the marked arc .

Thus, we see that inequality
satisfy the numbers for which the inequality is true
. We solved the inequality for numbers located on the same period of the sine function. Therefore, all solutions to the inequality can be written in the form

Students should be asked to carefully examine the drawing and figure out why all the solutions to the inequality
can be written in the form
,
.

Rice. 3

It is necessary to draw students' attention to the fact that when solving inequalities for the cosine function, we draw a straight line parallel to the ordinate axis.

Graphical method for solving inequalities.

We build graphs
And
, given that
.

Rice. 4

Then we write the equation
and his decision
,
,
, found using formulas
,
,
.

(Givingn values ​​0, 1, 2, we find the three roots of the compiled equation). Values
are three consecutive abscissas of the intersection points of the graphs
And
. Obviously, always on the interval
inequality holds
, and on the interval
– inequality
. We are interested in the first case, and then adding to the ends of this interval a number that is a multiple of the period of the sine, we obtain a solution to the inequality
as:
,
.

Rice. 5

Summarize. To solve the inequality
, you need to create the corresponding equation and solve it. Find the roots from the resulting formula And , and write the answer to the inequality in the form: ,
.

Thirdly, the fact about the set of roots of the corresponding trigonometric inequality is very clearly confirmed when solving it graphically.

Rice. 6

It is necessary to demonstrate to students that the turn, which is the solution to the inequality, is repeated through the same interval, equal to the period of the trigonometric function. You can also consider a similar illustration for the graph of the sine function.

Fourthly, it is advisable to carry out work on updating students’ techniques for converting the sum (difference) of trigonometric functions into a product, and to draw students’ attention to the role of these techniques in solving trigonometric inequalities.

Such work can be organized through students’ independent completion of tasks proposed by the teacher, among which we highlight the following:

Fifthly, students must be required to illustrate the solution to each simple trigonometric inequality using a graph or a trigonometric circle. You should definitely pay attention to its expediency, especially to the use of the circle, since when solving trigonometric inequalities, the corresponding illustration serves as a very convenient means of recording the set of solutions to a given inequality

It is advisable to introduce students to methods for solving trigonometric inequalities that are not the simplest ones according to the following scheme: turning to a specific trigonometric inequality turning to the corresponding trigonometric equation joint search (teacher - students) for a solution independent transfer of the found method to other inequalities of the same type.

In order to systematize students’ knowledge about trigonometry, we recommend specially selecting such inequalities, the solution of which requires various transformations that can be implemented in the process of solving it, and focusing students’ attention on their features.

As such productive inequalities we can propose, for example, the following:

In conclusion, we give an example of a set of problems for solving trigonometric inequalities.

1. Solve the inequalities:

2. Solve the inequalities: 3. Find all solutions to the inequalities: 4. Find all solutions to the inequalities:

A)
, satisfying the condition
;

b)
, satisfying the condition
.

5. Find all solutions to the inequalities:

A) ;

b) ;

V)
;

G)
;

d)
.

6. Solve the inequalities:

A) ;

b) ;

V) ;

G)
;

d) ;

e) ;

and)
.

7. Solve the inequalities:

A)
;

b) ;

V) ;

G) .

8. Solve the inequalities:

A) ;

b) ;

V) ;

G)
;

d)
;

e) ;

and)
;

h) .

It is advisable to offer tasks 6 and 7 to students studying mathematics at an advanced level, task 8 to students in classes with advanced study of mathematics.

§3. Special methods for solving trigonometric inequalities

Special methods for solving trigonometric equations - that is, those methods that can only be used to solve trigonometric equations. These methods are based on the use of the properties of trigonometric functions, as well as on the use of various trigonometric formulas and identities.

3.1. Sector method

Let's consider the sector method for solving trigonometric inequalities. Solving inequalities of the form

, WhereP ( x ) AndQ ( x ) – rational trigonometric functions (sines, cosines, tangents and cotangents are included in them rationally), similar to solving rational inequalities. It is convenient to solve rational inequalities using the method of intervals on the number line. Its analogue for solving rational trigonometric inequalities is the method of sectors in the trigonometric circle, forsinx Andcosx (
) or trigonometric semicircle fortgx Andctgx (
).

In the interval method, each linear factor of the numerator and denominator of the form
on the number axis corresponds to a point , and when passing through this point
changes sign. In the sector method, each factor of the form
, Where
- one of the functionssinx orcosx And
, in a trigonometric circle there correspond two angles And
, which divide the circle into two sectors. When passing through And function
changes sign.

The following must be remembered:

a) Factors of the form
And
, Where
, retain sign for all values . Such factors of the numerator and denominator are discarded by changing (if
) with each such rejection, the inequality sign is reversed.

b) Factors of the form
And
are also discarded. Moreover, if these are factors of the denominator, then inequalities of the form are added to the equivalent system of inequalities
And
. If these are factors of the numerator, then in the equivalent system of restrictions they correspond to the inequalities
And
in the case of a strict initial inequality, and equality
And
in the case of a non-strict initial inequality. When discarding the multiplier
or
the inequality sign is reversed.

Example 1. Solve inequalities: a)
, b)
. we have function b) . Solve the inequality We have,

3.2. Concentric circle method

This method is an analogue of the parallel number axes method for solving systems of rational inequalities.

Let's consider an example of a system of inequalities.

Example 5. Solve a system of simple trigonometric inequalities

First, we solve each inequality separately (Figure 5). In the upper right corner of the figure we will indicate for which argument the trigonometric circle is being considered.

Fig.5

Next, we build a system of concentric circles for the argumentX . We draw a circle and shade it according to the solution of the first inequality, then we draw a circle of a larger radius and shade it according to the solution of the second, then we construct a circle for the third inequality and a base circle. We draw rays from the center of the system through the ends of the arcs so that they intersect all the circles. We form a solution on the base circle (Figure 6).

Fig.6

Answer:
,
.

Conclusion

All objectives of the course research were completed. The theoretical material is systematized: the main types of trigonometric inequalities and the main methods for solving them are given (graphical, algebraic, method of intervals, sectors and the method of concentric circles). An example of solving an inequality was given for each method. The theoretical part was followed by the practical part. It contains a set of tasks for solving trigonometric inequalities.

This coursework can be used by students for independent work. Schoolchildren can check the level of mastery of this topic and practice completing tasks of varying complexity.

Having studied the relevant literature on this issue, we can obviously conclude that the ability and skills to solve trigonometric inequalities in the school course of algebra and elementary analysis are very important, the development of which requires significant effort on the part of the mathematics teacher.

Therefore, this work will be useful for mathematics teachers, as it makes it possible to effectively organize the training of students on the topic “Trigonometric inequalities.”

The research can be continued by expanding it to a final qualifying work.

List of used literature

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Annex 1

Graphic interpretation of solutions to simple inequalities

Rice. 1

Rice. 2

Fig.3

Fig.4

Fig.5

Fig.6

Fig.7

Fig.8

Appendix 2

Solutions to simple inequalities

During the 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 the specified value into it.

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) 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 us write the tangent of the 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 of not giving 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 solving a fractional equation that 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.

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, where 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, we will analyze the approach to solving the example without using formulas for 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 the right point, we, on the contrary, leave 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. A typical mistake 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 what follows we will use general formulas to solve 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, all real values ​​of the argument satisfy the inequality.

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

Problem 13. Solve inequality .

When solving inequalities containing trigonometric functions, they are reduced to the simplest inequalities of the form cos(t)>a, sint(t)=a and similar ones. And already the simplest inequalities are solved. Let's look at various examples of ways to solve simple trigonometric inequalities.

Example 1. Solve the inequality sin(t) > = -1/2.

Draw a unit circle. Since sin(t) by definition is the y coordinate, we mark the point y = -1/2 on the Oy axis. We draw a straight line through it parallel to the Ox axis. At the intersection of the straight line with the graph of the unit circle, mark the points Pt1 and Pt2. We connect the origin of coordinates with points Pt1 and Pt2 by two segments.

The solution to this inequality will be all points of the unit circle located above these points. In other words, the solution will be the arc l. Now it is necessary to indicate the conditions under which an arbitrary point will belong to the arc l.

Pt1 lies in the right semicircle, its ordinate is -1/2, then t1=arcsin(-1/2) = - pi/6. To describe point Pt1, you can write the following formula:
t2 = pi - arcsin(-1/2) = 7*pi/6. As a result, we obtain the following inequality for t:

We preserve the inequalities. And since the sine function is periodic, it means that the solutions will be repeated every 2*pi. We add this condition to the resulting inequality for t and write down the answer.

Answer: -pi/6+2*pi*n< = t < = 7*pi/6 + 2*pi*n, при любом целом n.

Example 2. Solve cos(t) inequality<1/2.

Let's draw a unit circle. Since, according to the definition, cos(t) is the x coordinate, we mark the point x = 1/2 on the graph on the Ox axis.
We draw a straight line through this point parallel to the Oy axis. At the intersection of the straight line with the graph of the unit circle, mark the points Pt1 and Pt2. We connect the origin of coordinates with points Pt1 and Pt2 by two segments.

The solutions will be all points of the unit circle that belong to the arc l. Let's find the points t1 and t2.

t1 = arccos(1/2) = pi/3.

t2 = 2*pi - arccos(1/2) = 2*pi-pi/3 = 5*pi/6.

We got the inequality for t: pi/3

Since cosine is a periodic function, the solutions will be repeated every 2*pi. We add this condition to the resulting inequality for t and write down the answer.

Answer: pi/3+2*pi*n

Example 3. Solve inequality tg(t)< = 1.

The tangent period is equal to pi. Let's find solutions that belong to the interval (-pi/2;pi/2) right semicircle. Next, using the periodicity of the tangent, we write down all the solutions to this inequality. Let's draw a unit circle and mark a line of tangents on it.

If t is a solution to the inequality, then the ordinate of the point T = tg(t) must be less than or equal to 1. The set of such points will make up the ray AT. The set of points Pt that will correspond to the points of this ray is the arc l. Moreover, point P(-pi/2) does not belong to this arc.

Most students don't like trigonometric inequalities. But in vain. As one character used to say,

“You just don’t know how to cook them”

So how to “cook” and with what to submit inequality with sine we will figure out in this article. We will solve it in the simplest way - using the unit circle.

So, first of all, we need the following algorithm.

Algorithm for solving inequalities with sine:

1. on the sine axis we plot the number $a$ and draw a straight line parallel to the cosine axis until it intersects with the circle;
2. the points of intersection of this line with the circle will be shaded if the inequality is not strict, and not shaded if the inequality is strict;
3. the solution area of ​​the inequality will be located above the line and up to the circle if the inequality contains the sign “$>$”, and below the line and up to the circle if the inequality contains the sign “$<$”;
4. to find the intersection points, we solve the trigonometric equation $\sin(x)=a$, we get $x=(-1)^(n)\arcsin(a) + \pi n$;
5. setting $n=0$, we find the first intersection point (it is located either in the first or fourth quarter);
6. to find the second point, we look in which direction we go through the area to the second intersection point: if in a positive direction, then we should take $n=1$, and if in a negative direction, then $n=-1$;
7. in response, the interval is written down from the smaller intersection point $+ 2\pi n$ to the larger one $+ 2\pi n$.

Algorithm limitation

Important: d given algorithm does not work for inequalities of the form $\sin(x) > 1; \ \sin(x) \geq 1, \ \sin(x)< -1, \ \sin{x} \leq -1$. В строгом случае эти неравенства не имеют решений, а в нестрогом – решение сводится к решению уравнения $\sin{x} = 1$ или $\sin{x} = -1$.

Special cases when solving inequalities with sine

It is also important to note the following cases, which are much more convenient to solve logically without using the above algorithm.

Special case 1. Solve inequality:

$\sin(x)\leq 1.$

Due to the fact that the range of values ​​of the trigonometric function $y=\sin(x)$ is not greater than modulo $1$, then the left side of the inequality at any$x$ from the domain of definition (and the domain of definition of the sine is all real numbers) is not more than $1$. And, therefore, in the answer we write: $x \in R$.

Consequence:

$\sin(x)\geq -1.$

Special case 2. Solve inequality:

$\sin(x)< 1.$

Applying reasoning similar to special case 1, we find that the left side of the inequality is less than $1$ for all $x \in R$, except for points that are solutions to the equation $\sin(x) = 1$. Solving this equation, we will have:

$x = (-1)^(n)\arcsin(1)+ \pi n = (-1)^(n)\frac(\pi)(2) + \pi n.$

And, therefore, in the answer we write: $x \in R \backslash \left\((-1)^(n)\frac(\pi)(2) + \pi n\right\)$.

Consequence: the inequality is solved similarly

$\sin(x) > -1.$

Examples of solving inequalities using an algorithm.

Example 1: Solve inequality:

$\sin(x) \geq \frac(1)(2).$

1. Let us mark the coordinate $\frac(1)(2)$ on the sine axis.
2. Let's draw a straight line parallel to the cosine axis and passing through this point.
3. Let's mark the intersection points. They will be shaded because the inequality is not strict.
4. The inequality sign is $\geq$, which means we paint the area above the line, i.e. smaller semicircle.
5. We find the first intersection point. To do this, we turn the inequality into equality and solve it: $\sin(x)=\frac(1)(2) \ \Rightarrow \ x=(-1)^(n)\arcsin(\frac(1)(2) )+\pi n =(-1)^(n)\frac(\pi)(6) + \pi n$. We further set $n=0$ and find the first intersection point: $x_(1)=\frac(\pi)(6)$.
6. We find the second point. Our area goes in the positive direction from the first point, which means we set $n$ equal to $1$: $x_(2)=(-1)^(1)\frac(\pi)(6) + \pi \cdot 1 = \ pi – \frac(\pi)(6) = \frac(5\pi)(6)$.

Thus, the solution will take the form:

$x \in \left[\frac(\pi)(6) + 2\pi n; \frac(5\pi)(6) + 2 \pi n\right], \n \in Z.$

Example 2: Solve inequality:

$\sin(x)< -\frac{1}{2}$

Let's mark the coordinate $-\frac(1)(2)$ on the sine axis and draw a straight line parallel to the cosine axis and passing through this point. Let's mark the intersection points. They will not be shaded, since the inequality is strict. The inequality sign $<$, а, значит, закрашиваем область ниже прямой, т.е. меньший полукруг. Неравенство превращаем в равенство и решаем его:

$\sin(x)=-\frac(1)(2)$

$x=(-1)^(n)\arcsin(\left(-\frac(1)(2)\right))+ \pi n =(-1)^(n+1)\frac(\pi )(6) + \pi n$.

Further assuming $n=0$, we find the first intersection point: $x_(1)=-\frac(\pi)(6)$. Our area goes in the negative direction from the first point, which means we set $n$ equal to $-1$: $x_(2)=(-1)^(-1+1)\frac(\pi)(6) + \pi \cdot (-1) = -\pi + \frac(\pi)(6) = -\frac(5\pi)(6)$.

So, the solution to this inequality will be the interval:

$x \in \left(-\frac(5\pi)(6) + 2\pi n; -\frac(\pi)(6) + 2 \pi n\right), \n \in Z.$

Example 3: Solve inequality:

$1 – 2\sin(\left(\frac(x)(4)+\frac(\pi)(6)\right)) \leq 0.$

This example cannot be solved immediately using an algorithm. First you need to transform it. We do exactly what we would do with an equation, but don’t forget about the sign. Dividing or multiplying by a negative number reverses it!

So, let's move everything that does not contain a trigonometric function to the right side. We get:

$- 2\sin(\left(\frac(x)(4)+\frac(\pi)(6)\right)) \leq -1.$

Let's divide the left and right sides by $-2$ (don't forget about the sign!). Will have:

$\sin(\left(\frac(x)(4)+\frac(\pi)(6)\right)) \geq \frac(1)(2).$

Again we have an inequality that we cannot solve using an algorithm. But here it is enough to change the variable:

$t=\frac(x)(4)+\frac(\pi)(6).$

We obtain a trigonometric inequality that can be solved using the algorithm:

$\sin(t) \geq \frac(1)(2).$

This inequality was solved in Example 1, so let's borrow the answer from there:

$t \in \left[\frac(\pi)(6) + 2\pi n; \frac(5\pi)(6) + 2 \pi n\right].$

However, the decision is not over yet. We need to go back to the original variable.

$(\frac(x)(4)+\frac(\pi)(6)) \in \left[\frac(\pi)(6) + 2\pi n; \frac(5\pi)(6) + 2 \pi n\right].$

Let's imagine the interval as a system:

$\left\(\begin(array)(c) \frac(x)(4)+\frac(\pi)(6) \geq \frac(\pi)(6) + 2\pi n, \\ \frac(x)(4)+\frac(\pi)(6) \leq \frac(5\pi)(6) + 2 \pi n. \end(array) \right.$

On the left side of the system there is an expression ($\frac(x)(4)+\frac(\pi)(6)$), which belongs to the interval. The left boundary of the interval is responsible for the first inequality, and the right boundary is responsible for the second. Moreover, brackets play an important role: if the bracket is square, then the inequality will be relaxed, and if it is round, then it will be strict. our task is to get $x$ on the left in both inequalities.

Let's move $\frac(\pi)(6)$ from the left side to the right side, we get:

$\left\(\begin(array)(c) \frac(x)(4) \geq \frac(\pi)(6) + 2\pi n -\frac(\pi)(6), \\ \frac(x)(4) \leq \frac(5\pi)(6) + 2 \pi n – \frac(\pi)(6).\end(array) \right.$

Simplifying, we will have:

$\left\(\begin(array)(c) \frac(x)(4) \geq 2\pi n, \\ \frac(x)(4) \leq \frac(2\pi)(3) + 2 \pi n. \end(array) \right.$

Multiplying the left and right sides by $4$, we get:

$\left\(\begin(array)(c) x \geq 8\pi n, \\ x \leq \frac(8\pi)(3) + 8 \pi n. \end(array) \right. $

Assembling the system into the interval, we get the answer:

$x \in \left[ 8\pi n; \frac(8\pi)(3) + 8 \pi n\right], \n \in Z.$

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