How to solve trigonometric inequalities on a circle. Solving trigonometric inequalities

An algorithm for solving simple trigonometric inequalities and recognizing methods for solving trigonometric inequalities.

Teachers of the highest qualification category:

Shirko F.M. p. Progress, MOBU-secondary school No. 6

Sankina L.S. Armavir, private secondary school "New Way"

There are no universal methods for teaching science and mathematics disciplines. Each teacher finds his own ways of teaching that are acceptable only to him.

Our many years of teaching experience show that students more easily learn material that requires concentration and retention of a large amount of information in memory if they are taught to use algorithms in their activities at the initial stage of learning a complex topic. In our opinion, such a topic is the topic of solving trigonometric inequalities.

So, before we begin with students to identify techniques and methods for solving trigonometric inequalities, we practice and consolidate an algorithm for solving the simplest trigonometric inequalities.

Algorithm for solving simple trigonometric inequalities

Mark points on the corresponding axis ( For sin x– OA axis, forcos x– OX axis)

We restore a perpendicular to the axis that will intersect the circle at two points.

The first point on the circle is a point that belongs to the interval of the arc function range by definition.

Starting from the labeled point, shade the arc of the circle corresponding to the shaded part of the axis.

We pay special attention to the direction of the detour. If the traversal is done clockwise (i.e. there is a transition through 0), then the second point on the circle will be negative, if counterclockwise it will be positive.

We write the answer in the form of an interval, taking into account the periodicity of the function.

Let's look at the operation of the algorithm using examples.

1) sin ≥ 1/2;

Solution:

We depict a unit circle.;

We mark point ½ on the OU axis.

We restore the perpendicular to the axis,

which intersects the circle at two points.

By definition of the arcsine, we first note

point π/6.

Shade the part of the axis that corresponds to

given inequality, above the point ½.

Shade the arc of the circle corresponding to the shaded part of the axis.

The traversal is done counterclockwise, we get the point 5π/6.

We write the answer in the form of an interval, taking into account the periodicity of the function;

Answer:x;[π/6 + 2π n, 5π/6 + 2π n], n Z.

The simplest inequality is solved using the same algorithm if the answer record does not contain a table value.

Students, when solving inequalities at the board in their first lessons, recite each step of the algorithm out loud.

2) 5 cos x – 1 ≥ 0;

R solution:at

5 cos x – 1 ≥ 0;

cos x ≥ 1/5;

Draw a unit circle.

We mark a point with coordinate 1/5 on the OX axis.

We restore the perpendicular to the axis, which

intersects the circle at two points.

The first point on the circle is a point that belongs to the interval of the arc cosine range by definition (0;π).

We shade the part of the axis that corresponds to this inequality.

Starting from the signed point arccos 1/5, shade the arc of the circle corresponding to the shaded part of the axis.

The traversal is done clockwise (i.e. there is a transition through 0), which means that the second point on the circle will be negative - arccos 1/5.

We write the answer in the form of an interval, taking into account the periodicity of the function, from the smaller value to the larger one.

Answer: x  [-arccos 1/5 + 2π n, arccos 1/5 + 2π n], n Z.

Improving the ability to solve trigonometric inequalities is facilitated by the following questions: “How will we solve a group of inequalities?”; “How does one inequality differ from another?”; “How is one inequality similar to another?”; How would the answer change if strict inequality were given?"; How would the answer change if instead of the sign "" there was a sign "

The task of analyzing a list of inequalities from the standpoint of methods for solving them allows you to practice their recognition.

Students are given inequalities that need to be solved in class.

Question: Highlight the inequalities that require the use of equivalent transformations when reducing a trigonometric inequality to its simplest form?

Answer 1, 3, 5.

Question: What are the inequalities in which you need to consider a complex argument as a simple one?

Answer: 1, 2, 3, 5, 6.

Question: What are the inequalities where trigonometric formulas can be applied?

Answer: 2, 3, 6.

Question: Name the inequalities where the method of introducing a new variable can be applied?

Answer: 6.

The task of analyzing a list of inequalities from the standpoint of methods for solving them allows you to practice their recognition. When developing skills, it is important to identify the stages of its implementation and formulate them in a general form, which is presented in the algorithm for solving the simplest 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, it becomes completely unnecessary to carefully solve a large number of exercises under the guidance of a teacher in order to demonstrate all sorts of reasoning techniques depending on the sign of the 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...

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 arguments 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 $x$ from 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 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:

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

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 indicated with a “+” sign.

Fig.7

Final answer:

Note. If a wavy line, after traversing 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. The 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 pass 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:

A)
, satisfying the condition
;

b)
, satisfying the condition
.

5. Find all solutions to the inequalities:

A) ;

b) ;

V)
;

G)
;

e)
.

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)
;

e)
;

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 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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Zhurbenko, L.N. Mathematics in examples and problems [Text] / L.N. Zhurbenko. – M.: Infra-M, 2009. – 373 p.

Ivanov, O.A. Elementary mathematics for schoolchildren, students and teachers [Text] / O.A. Ivanov. – M.: MTsNMO, 2009. – 384 p.

Karp, A.P. Assignments on algebra and the beginnings of analysis for organizing final repetition and certification in grade 11 [Text] / A.P. Carp. – M.: Education, 2005. – 79 p.

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Annex 1

Graphic interpretation of solutions to simple inequalities

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Fig.3

Fig.4

Fig.5

Fig.6

Fig.7

Fig.8

Appendix 2

Solutions to simple inequalities

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. Make a reverse substitution (return to the original argument) and express the value X from the double inequality, we 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 straight. 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 on the UNT or Unified State Examination when solving a trigonometric inequality of the form cost