Ivanova Anastasia

Task No. 15 of the specialized exam in mathematics is a task of an increased level of complexity, representing inequality. When solving these inequalities, students must demonstrate knowledge of theorems on the equivalence of inequalities of a certain type, and the ability to use standard and non-standard solution methods. An analysis of the content of school textbooks shows that in most of them, methods for solving inequalities using the properties of functions are not given due attention, and in the Unified State Examination tasks almost every year inequalities are proposed, the solution of which is simplified if the properties of functions are applied. According to statistics presented on the website of the Federal Institute of Pedagogical Measurements, in 2017, about 15% of exam participants received non-zero points for this task; the maximum score is about 11%. Everything noted indicates that students experience great difficulties in solving task No. 15 of the Unified State Exam. Target: Explore different ways to solve inequalities.

:

1. Study theoretical material on this topic.

2. Consider the examples offered in the Unified State Exam task bank on the website of the Federal Institute of Pedagogical Measurements.

3. Study functional-graphical methods for solving inequalities.

4. Compare different methods for solving inequalities.

5. Check experimentally which method of solving inequalities is the most rational.

Research methods: survey, questioning, analysis, comparison and synthesis of results.

In our work, we studied functional-graphical methods for solving inequalities. Various methods for solving inequalities were compared. We checked experimentally which method of solving inequalities is the most rational. And they came to the conclusion that the student must know several ways to solve inequalities in order to save time and reduce the risk of logical and computational errors.

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Study of various methods for solving inequalities

Ivanova Anastasia Evgenevna

Municipal budgetary educational institution
"Secondary school No. 30 with in-depth study of individual subjects"

11b grade

Scientific article (job description)

1. Introduction

Relevance.

Task No. 15 of the specialized exam in mathematics is a task of an increased level of complexity, representing inequality (rational, irrational, exponential, logarithmic). When solving these inequalities, students must demonstrate knowledge of theorems on the equivalence of inequalities of a certain type, and the ability to use standard and non-standard solution methods.

A complete correct solution to this task is worth 2 points. When solving a problem, any mathematical methods are acceptable - algebraic, functional, graphical, geometric, etc.

According to statistics presented on the website of the Federal Institute of Pedagogical Measurements, in 2017, about 15% of exam participants received non-zero points for this task; the maximum score is about 11%. Typical errors are associated with inattentive reading of the mathematical notation of inequalities, misunderstanding of the algorithm for solving aggregates and systems of logarithmic inequalities. A lot of mistakes were made by exam participants when solving fractional rational inequalities (the denominator was forgotten).

The results of completing task No. 15 by students of our school on the Unified State Exam in mathematics are presented in Table 1 and in the diagram (Fig. 1).

Table 1

Results of task No. 15 by students of our school

Fig.1. Results of task No. 15 by students of our school

The results of task No. 15 on the trial city exam of grades 11a,b in the 2017-2018 school year. year are presented in Table 2 and in the diagram (Fig. 2).

table 2

Results of task No. 15 on the trial city exam

in the 2017-2018 academic year. year by students of our school

Fig.2. The results of task No. 15 on the trial exam in the 2017-2018 academic year. year by students of our school

We conducted a survey of mathematics teachers at our school and identified the main problems that students have when solving inequalities: incorrect determination of the range of acceptable values ​​of inequalities; consideration of not all cases of transition from logarithmic inequality to rational; converting logarithmic expressions; errors in using the interval method, etc.

A number of typical errors are associated with the use of the interval method and the introduction of an auxiliary variable. For example, an error in determining signs on intervals or incorrect placement of numbers on a coordinate line, according to the criteria, can be interpreted as computational errors. Others related to skipping steps of the algorithm or performing them incorrectly are scored 0 points.

Everything noted indicates that students experience great difficulties in solving task No. 15 of the Unified State Exam in mathematics. In this regard, we have put forward hypothesis : if a student knows several ways to solve inequalities, then he will be able to choose the most rational one.

Object of study: inequalities.

Subject of study: various ways to solve inequalities.

Target : Explore different ways to solve inequalities.

To achieve this goal, the following tasks were solved:

1. Study theoretical material on this topic.
2. Consider the examples offered in the Unified State Exam task bank on the website of the Federal Institute of Pedagogical Measurements.
3. Study functional-graphical methods for solving inequalities.
4. Compare different methods for solving inequalities.
5. Check experimentally which method of solving inequalities is the most rational.

2. Main part

2.1. Theoretical part

1. Linear inequalities

Linear inequalitiesare inequalities of the form: ax + b 0; ax+b≥0; ax+b≤0, where a and b – any numbers, and a≠0,x - unknown variable.

Rules for transforming inequalities:

1. Any term of the inequality can be transferred from one part of the inequality to another, changing the sign to the opposite one.

2. Both sides of the inequality can be multiplied/divided by the same positive number to obtain an inequality equivalent to the given one.

3. Both sides of the inequality can be multiplied/divided by the same negative number, reversing the sign of the inequality.

2. Quadratic inequalities

Inequality of the form

where x is a variable, a, b, c are numbers, , is called square. When solving a quadratic inequality, it is necessary to find the roots of the correspondingquadratic equation . To do this you need to finddiscriminant of this quadratic equation. You can get 3 cases: 1) D=0 , a quadratic equation has one root; 2) D>0 a quadratic equation has two roots; 3)D a quadratic equation has no roots. Depending on the obtained roots and the sign of the coefficient a one of six locations possiblefunction graphics (Annex 1).

3. Rational inequalities

Rational inequalitywith one variable x is called an inequality of the form f(x) expressions, i.e. algebraic expressions made up of numbers, the variable x and using mathematical operations, i.e. operations of addition, subtraction, multiplication, division and raising to natural powers.Algorithm for solving rational inequalities using the interval method(Annex 1).

4. Exponential inequalities

Exponential inequality- this is inequality , in which the unknown is in the exponent. The simplestexponential inequality has the form: a x ‹ b or a x › b, where a> 0, a ≠ 1, x is unknown.

5. Logarithmic inequalities

Logarithmic inequalitycalled an inequality in which the unknown quantity is under the signlogarithm .

1. Inequality if reduces to equivalent inequality. If - then to inequality.

Similarly the inequalityis equivalent to the inequalities for: ; For : .

The solutions to the obtained inequalities must be intersected with the ODZ:

2. Solving a logarithmic inequality of the formis equivalent to solving the following systems:

A) b)

Inequality in each of the two cases it is reduced to one of the systems:

A) b)

6. Irrational inequalities

If the inequality includes functions under the root sign, then such inequalities are called irrational.

.

2.2. Practical part

Study #1

Target : Learn the limited function method.

Progress:

1. Study the method of limited functions.

2. Solve inequalities by this method.

To use the boundedness of a function, you must be able to find the set of values ​​of a function and know estimates of the range of values ​​of standard functions (for example,) .

Example #1 . Solve inequality:

Solution:

Domain:

For all x from the resulting set we have:

Therefore, the solution to the inequality

Answer:

Example No. 2. Solve inequality:

Solution:

Because

This inequality is equivalent

The first equation of the system has one root x = - 0.4, which also satisfies the second equation.

Answer: - 0.4

Conclusion: This method is most effective if the inequality contains functions such asand others, the ranges of which are limited above or below.

Study #2

Target : study the method of rationalizing the solution of inequalities.

Progress:

1. Study the rationalization method.

2. Solve inequalities by this method.

The rationalization method consists of replacing the complex expression F(x) with a simpler expression G(x), in which the inequality G(x) v 0 is equivalent to the inequality F(x) v 0 on the domain of definition of the expression F(x) (symbol "v" replaces one of the inequality signs: ≤, ≥, >,

Let us highlight some typical expressions F and the corresponding rationalizing expressions G (Table 1), where f, g, h, p, q are expressions with variable x (h>0, h≠1,f>0,g>0), a -fixed number (a>0, a≠1). (Appendix 2).

Example No. 1. Solve inequality:

O.D.Z:

Answer:

Example No. 2. Solve inequality:

O.D.Z:

Taking into account the domain of definition, we obtain

Answer:

Conclusion : Inequalities with logarithms to a variable base are the most difficult. The rationalization method allows you to move from inequalities containing complex exponential, logarithmic, etc. expression, to its equivalent simpler rational inequality. The rationalization method not only saves time, but also reduces the risk of logical and computational errors.

Study #3

Target : in the process of solving inequalities, compare different methods.

Progress:

1. Solve inequalityusing different methods.

2. Compare the results and draw a conclusion.

Example No. 1. Solve inequality

Solution:

1 way. Algebraic method

Solution of the first system:

We solve the second inequality of the second system:

Method 2 . Using Function Scope

Domain:

For these x values ​​we get:

The right side of the inequality is negative on its domain of definition. Therefore, the inequality is valid for

Answer:

3 way. Graphical method

Conclusion : while solving the inequality using the algebraic method, I came to an inequality of the sixth degree, spent a lot of time solving it, but could not solve it. A rational method, in my opinion, is to use the function domain or graphically.

Example No. 2. Solve inequality:.

Answer:

Conclusion: I was able to solve this inequality only thanks to the rationalization method.

Conclusion

An analysis of the content of school textbooks shows that in most of them, methods for solving inequalities using the properties of functions are not given due attention, and in the Unified State Examination tasks almost every year inequalities are proposed, the solution of which is simplified if the properties of functions are applied.

Most students solve inequalities using standard, algorithmic methods, which sometimes results in cumbersome calculations. In this regard, the percentage of completion of task No. 15 on the Unified State Exam is low.

The scope of application of the properties of functions in solving inequalities is very wide. The use of properties (boundedness, monotonicity, etc.) of the functions included in the inequalities allows the use of non-standard solution methods. In our opinion, the ability to use the necessary properties of functions when solving inequalities can allow students to choose a more rational solution.

In our work, we studied functional-graphical methods for solving inequalities. Various methods for solving inequalities were compared. We checked experimentally which method of solving inequalities is the most rational.

And they came to the conclusion that the student must know several ways to solve inequalities in order to save time and reduce the risk of logical and computational errors.

The objectives of our work have been completed, the goal has been achieved, the hypothesis has been confirmed.

Literature:

1. Alimov Sh. A, Kolyagin Yu. M., Sidorov Yu. V. et al. Algebra and the beginnings of analysis: Textbook for grades 10-11. general education institutions / Sh. A. Alimov, Yu. M. Kolyagin, Yu. V. Sidorov and others - 15th ed. – M.: Education, 2007. – 384 p.
2. Koryanov A.G., Prokofiev A.A. Materials of the course “Preparing good and excellent students for the Unified State Exam”: lectures 1-4. - M.: Pedagogical University “First of September”, 2012. – 104 p.
3. Website http://www.fipi.ru/.
4. Website https://ege.sdamgia.ru/.
5. Yashchenko I. V. Unified State Exam. Mathematics. Profile level: standard exam options: 36 options / ed. I. V. Yashchenko. - M.: Publishing House "National Education", 2018. - 256 p.

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Slide captions:

Study of various methods for solving inequalities Ivanova Anastasia Evgenievna MBOU "Secondary school No. 30 with in-depth study of individual subjects"

Results of task No. 15 by students of our school

The results of task No. 15 on the trial exam in the 2017-2018 academic year. year by students of our school

Hypothesis: if a student knows several ways to solve inequalities, then he will be able to choose the most rational Object of study: inequalities Subject of study: various ways to solve inequalities

Goal: Explore different ways to solve inequalities. To achieve this goal, the following tasks were solved: Study theoretical material on this topic. Consider the examples offered in the Unified State Exam task bank on the website of the Federal Institute of Pedagogical Measurements. Study functional-graphical methods for solving inequalities. Compare different methods for solving inequalities. Check experimentally which method of solving inequalities is the most rational.

Study No. 1 Purpose: To study the limited function method. Progress: 1. Study the method of limited functions. 2. Solve inequalities using this method. Example No. 1. Solve the inequality: Solution: Domain: For all x from the resulting set we have: Therefore, the solution to the inequality Answer:

Example No. 2. Solve the inequality: Solution: Because This inequality is equivalent. The first equation of the system has one root x = - 0.4, which also satisfies the second equation. Answer: - 0.4 Conclusion: this method is most effective if the inequality contains functions, like others, whose ranges of values ​​are limited above or below.

Study No. 2 Purpose: to study a method for rationalizing the solution of inequalities. Progress: 1. Study the rationalization method. 2. Solve inequalities using this method. Example No. 1. Solve the inequality: O.D.Z: Taking into account the domain of definition, we obtain Answer:

Example No. 2. Solve the inequality: O.D.Z: Taking into account the domain of definition, we get Answer: Conclusion: inequalities with logarithms to a variable base cause the greatest difficulty. The rationalization method allows you to move from inequalities containing complex exponential, logarithmic, etc. expression, to its equivalent simpler rational inequality. The rationalization method not only saves time, but also reduces the risk of logical and computational errors.

Study No. 3 Purpose: in the process of solving inequalities, compare different methods. Progress: 1. Solve the inequality using different methods. 2. Compare the results and draw a conclusion. Example No. 1. Solve inequality 1 way. Algebraic method Solving the first system: Solving the second inequality of the second system: 2nd method. Using the domain of definition of a function Domain of definition: For these values ​​of x we ​​get: The right side of the inequality is negative on its domain of definition. Therefore, the inequality is valid for

3 way. Graphical method Conclusion: solving the inequality using the algebraic method, I came to an inequality of the sixth degree, spent a lot of time solving it, but could not solve it. A rational method, in my opinion, is to use the function domain or graphically.

Example No. 2. Solve the inequality: Answer: Conclusion: I was able to solve this inequality only thanks to the rationalization method.

The scope of application of the properties of functions in solving inequalities is very wide. The use of properties (boundedness, monotonicity, etc.) of the functions included in the inequalities allows the use of non-standard solution methods. In our opinion, the ability to use the necessary properties of functions when solving inequalities can allow students to choose a more rational solution. In our work, we studied functional-graphical methods for solving inequalities. Various methods for solving inequalities were compared. We checked experimentally which method of solving inequalities is the most rational. And they came to the conclusion that the student must know several ways to solve inequalities in order to save time and reduce the risk of logical and computational errors. The objectives of our work have been completed, the goal has been achieved, the hypothesis has been confirmed.

Thank you for your attention!

Municipal educational institution

Yuryevskaya basic secondary school

Ostrovsky district

Municipal stage of the regional methodological competition

Nomination

Toolkit

Subject

Functional-graphical method for solving equations and inequalities in a high school algebra course.

The work was prepared by:

mathematic teacher

Introduction

Analysis of school textbooks

Unified State Exam Analysis

1. General theoretical part

1.1. Graphical method

1.2. Functional method

2. Solving equations and inequalities using the properties of input

functions in them

2.1. Use of DZ

2.2. Using feature limitations

2.3. Using function monotonicity

2.4. Using Function Graphs

2.5. Using the even or odd properties and periodicity of functions .

3. Solving equations and inequalities

3.1. Solving equations

3.2. Solving inequalities

Workshop

Bibliography

Application

Introduction

The topic of my work is “Functional-graphical method for solving equations and inequalities in a high school algebra course.” One of the main topics of the high school algebra course. Solving equations and inequalities play an important role in high school mathematics courses. Students begin to learn about inequalities and equations in elementary school.

The content of the topics “Equations” and “Inequalities” is gradually deepened and expanded. So, for example, the percentage of inequalities from the total material studied in 7th grade is 20%, in 8th grade - 25%, in 9th grade - 30%, in 10-11th grades - 35%.

The final study of inequalities and equations occurs in the algebra and beginning analysis courses of grades 10-11. Some universities include equations and inequalities in exam papers, which are often very complex and require different approaches to solution. At school, one of the most difficult sections of the school mathematics course is covered only in a few elective classes.

The focus of this work is to provide a more complete disclosure of the application of the functional graphical method to solving equations and inequalities in the high school algebra course.

The relevance of this work is that this topic is included in the Unified State Exam.

In preparing this work, I set a goal to consider as many types of equations and inequalities as possible, solved by the functional-graphical method. Also study this topic more deeply, identifying the most rational solution that quickly leads to an answer.

The object of the study is algebra for grades 10-11, edited and variants of the Unified State Examination.

This work discusses frequently encountered types of equations and inequalities; I hope that the knowledge I gained in the process of work will help when passing school exams and when entering a university. It can also serve as a teaching aid for preparing schoolchildren to take the Unified State Exam.

Analysis of school textbooks

In the methodological literature, it is customary to divide all the methods on which the school line of equations and inequalities from grades 7 to 11 are divided into three groups:

factorization method;

ümethod of introducing new variables;

Functional graphic method.

Let's consider the third method, namely, the use of function graphs and various properties of functions.

Schoolchildren must be taught to use the functional-graphic method from the very beginning of studying the topic “Equations”.

The solution to some problems can be based on the properties of monotonicity, periodicity, evenness or oddness, etc. of the functions included in them.

Having analyzed the textbooks, we can conclude that this topic is discussed only in mathematics textbooks of the new generation. The construction of the course in these textbooks is based on the priority of the functional-graphic line. In other textbooks, the functional-graphical method of solving equations and inequalities is not highlighted as a separate topic. Using function properties to solve problems is mentioned in passing when studying other topics. The new textbooks also contain a sufficient number of tasks of this type. The textbook contains advanced level tasks. The most complete system of tasks is presented, systematized for each property of the function.

Textbook	“Algebra and beginnings of analysis 10-11”, textbook for educational institutions,	, “Algebra and beginnings of analysis 11”, textbook for general education institutions (profile level)		and others. “Algebra and the beginnings of analysis 11”, textbook for educational institutions	and others. “Algebra and the beginnings of analysis 10-11”, textbook for educational institutions
Place in the know	Chapter 8 “Equations and inequalities. Systems of equations and inequalities" (last topic of the course)	Chapter 6 “Equations and inequalities. Systems of equations and inequalities" (last topic of the course)	Chapter II “Equations, inequalities, systems”	There is no separate topic. But in the topic “Solving trigonometric equations and inequalities” a root theorem is formulated, which is used in further study	No separate topic
	§ §56 General methods for solving equations and inequalities (functional-graphical method: root theorem, boundedness of a function)	§ §27 General methods for solving equations and inequalities (functional-graphical method: root theorem, boundedness of a function)	§ Equations (inequalities) of the form ; § §12*Non-standard methods for solving equations and inequalities (using domains of existence of functions, non-negativity of functions, boundedness, using the sin and cos properties, using the derivative)	The property of monotonicity of a function, even-odd (when deriving formulas for the roots of trigonometric equations)	The property of monotonicity is mentioned when analyzing an example in the topic “Exponential Function”
Examples of considered equations and inequalities	(;		Solve the equation. How many roots does the equation have in this interval?	Solve the equation

Analysis of the Unified State Examination (texts and results)

The Unified State Exam as a form of certification, which was introduced into the practice of Russian education in 2002, has been moving from experimental to regular mode since 2009.

An analysis of the Unified State Exam texts showed that tasks in which the properties of functions are used are encountered every year.

In 2003, in tasks A9 and C2, when solving, you can apply the properties of functions:

· A9. Indicate the interval to which the roots of the equation belong .

· C2. Find all values p, for which the equation has no roots.

· In 2004 – task B2. How many roots does the equation have? .

· In 2005, task C2 (solve the equation ) completed by 37% of students.

In 2007, when completing the task “Solve the equation” in part B, graduates considered two cases when solving the equation, habitually revealing the modulus sign..gif" width="81" height="24"> takes only positive values.

Even well-prepared students often complete tasks using “cookie-cutter” solution methods that lead to cumbersome transformations and calculations.

Obviously, when completing the above tasks, a well-prepared graduate had to show not only knowledge of known methods for solving equations or transforming expressions, but also the ability to analyze the condition, correlate the data and requirements of the task, derive various consequences from the condition, etc., that is show a certain level of development of mathematical thinking.

Thus, when teaching well-performing students, you need to not only take care of mastering the basic component of the algebra course and the beginnings of analysis (mastering the learned rules, formulas, methods), but also about realizing one of the main goals of teaching mathematics - the development of students’ thinking, in particular, mathematical thinking. Elective courses can serve to achieve this goal.

Indeed, when studying mathematics, students of general education institutions are traditionally introduced to the graphical method of solving equations, inequalities and their systems. However, in recent years, new classes of equations (inequalities) and new functional methods for solving them have appeared in the content of mathematics teaching. However, the tasks contained in the test materials of the Unified State Exam (USE) (the so-called combined equations), the solutions of which require the use of only the functional-graphical method, cause difficulties for students.

1. General theoretical part

Let X and Y be two arbitrary numerical sets. The elements of these sets will be denoted by x and y, respectively, and will be called variables.

Definition. A numerical function defined on the set X and taking values ​​in the set Y is called a correspondence (rule, law) that associates each x from the set X with one and only one value y from the set Y.

The variable x is called the independent variable or argument, and the variable y is the dependent variable. It is also said that the variable y is function from the variable x. The values ​​of the dependent variable are called the values ​​of the function.

The introduced concept of a numerical function is a special case of the general concept of a function as a correspondence between elements of two or more arbitrary sets.

Let X and Y be two arbitrary sets.

Definition. A function defined on the set X and taking values ​​in the set Y is a correspondence that associates with each element of the set X one and only one element from the set Y.

Definition. To define a function means to indicate the scope of its definition and the correspondence (rule) with the help of which, given the value of the independent variable, the corresponding function values ​​are found.

There are two ways to solve equations associated with the concept of function: graphic And functional. A special case of a functional method is the method functional, or universal substitutions.

Definition. Solving a given equation means finding the set of all its roots (solutions). The set of roots (solutions) can be empty, finite or infinite. In the following chapters of the theoretical section, we will analyze the above-described methods for solving equations, and in the “Practice” section we will show their application in various situations.

1.1. Graphic method.

In practice, to construct a graph of some functions, they compile a table of function values ​​for some argument values, then plot the corresponding points on the coordinate plane and sequentially connect them with a line. It is assumed that the points sufficiently accurately show the progress of the change in the function.

Definition. The graph of a function y = f(x) is the set of all points

(x, f(x) | x https://pandia.ru/text/78/500/images/image024_0.jpg" width="616" height="403">

The intersection point of the graphs has coordinates (0.5; 0). Hence, x=0.5

Answer: x=0.5

Example 2.

10| sinx|=10|cosx|-1

This equation can be rationally solved using the graphic-analytical method.

Since 10>1, this equation is equivalent to the following:

The intersection points of the graphs have coordinates ();. Therefore x=.

Answer: x=

1.2. Functional method

Not every equation of the form f(x)=g(x) as a result of transformations can be reduced to an equation of one or another standard form, for which conventional solution methods are suitable. In such cases, it makes sense to use such properties of the functions f(x) and g(x) as monotonicity, boundedness, parity, periodicity, etc. So, if one of the functions increases and the other decreases over a certain interval, then the equation f(x) = g(x) cannot have more than one root, which, in principle, can be found by selection. Further, if the function f(x) is bounded above and the function g(x) is bounded below so that f(x) swing=A g(x) min=A, then the equation f(x)=g(x) is equivalent to the system of equations

Also, when using the functional method, it is rational to use some of the theorems given below. To prove and use them, the following general equations are needed:

(2)

Theorem 1. The roots of equation (1) are the roots of equation (2).

Theorem 2. If f(x) is an increasing function on the interval a

The last theorem yields a corollary that is also used in solutions:

Corollary 1. If f(x) increases throughout its entire domain of definition, then on a given interval equations (1) and (2) are equivalent. If f(x) decreases over its entire domain of definition, n is odd, then on a given interval equations (1) and (2) are equivalent.

Theorem 3. If in the equation f(x)=g(x) for any admissible x the conditions f(x)≥a, g(x)≤a are satisfied, where a is some real number, then the given equation is equivalent to the system

Corollary 2. If in the equation f(x)+g(x)=a+b for any admissible x f(x)≤a, g(x)≤b, then this equation is equivalent to the system

The functional method for solving equations is often used in combination with the graphical one, since both of these methods are based on the same properties of functions. Sometimes a combination of these methods is called graphic-analytical method.

Example 1.

coshttps://pandia.ru/text/78/500/images/image033_3.gif" width="64" height="41 src=">≤1 x2+1≥1 =>

coshttps://pandia.ru/text/78/500/images/image035_3.gif" width="121" height="48">

=> x=π, at k=0

Answer: x=π

1.3. Functional substitution method

A special case of the functional method is the method of functional substitution - perhaps the most common method for solving complex problems in mathematics. The essence of the method is to introduce a new variable y=ƒ(x), the use of which leads to a simpler expression. A separate case of functional substitution is trigonometric substitution.

Trigonometric equation of the form

R(sin kx,cos nx, tg mx,ctg lx) = 0 (3)

where R is a rational function, k,n,m,lОZ, using trigonometric formulas for double and triple arguments, as well as addition formulas, can be reduced to a rational equation for the arguments sin x,cos x, tg x,ctg x, after which equation (3) can be reduced to a rational equation for t=tg( x/2) using universal trigonometric substitution formulas

2tg(x/2) 1-tg²(x/2)

Sin x=cos x=

1+tg²(x/2) 1+tg²(x/2)

2tg(x/2) 1-tg²(x/2)

Tg x=ctg x=

1-tg²(x/2) 2tg(x/2)

It should be noted that the use of formulas (4) can lead to a narrowing of the OD of the original equation, since tan(x/2) is not defined at the points x=π+2πk, kÎZ, therefore in such cases it is necessary to check whether the angles x=π+ 2πk, kÎZ roots of the original equation.

Example 1.

sin x+√2-sin² x+ sin x√2-sin² x = 3

Let now r = u+v and s=uv, then from the system of equations it follows

Since u = sin x and u = 1, then sin x= 1 and x = π/2+2πk, kО Z

Answer: x = π/2+2πk, kОZ

Example 2.

5 sin x-5 tg x

+4(1- cos x)=0

sin x+ tg x

This equation can be rationally solved using the functional substitution method.

Since tg x not defined at x = π/2+πk, kО Z, and sin x+tg x=0 at x = πk, kО Z, then the angles x = πk/2, kО Z are not included in the ODZ equations.

We use the formulas for the tangent of a half angle and denote t=tg( x/2), and according to the conditions of the problem t≠0;±1, then we get

https://pandia.ru/text/78/500/images/image055_2.gif" width="165"> +4 1- =0

Since t≠0;±1, this equation is equivalent to the equation

5t² + = 0 ó-5-5t² + 8 = 0

whence t = ± ..gif" width="27" height="47"> +2πk, kÎ Z

Example 3.

tg x+ ctg x+ tg²x+ ctg²x+ tg³x+ ctg³x=6

This equation can be rationally solved using the functional substitution method.

Let y=tg x+ctg x, then tg² x+ctg² x=y²-2, tg³ x+ctg³ x=y³-3y

Since tg x+ctg x=2, then tg x+1/ tg x=2. It follows that tg x=1 and x = π/4+πk, kО Z

Answer: x = π/2+2πk, kО Z

2. Solving equations and inequalities using the properties of the functions included in them

2. 1. Use of ODZ.

Sometimes knowledge of the ODZ allows you to prove that an equation (or inequality) has no solutions, and sometimes it allows you to find solutions to the equation (or inequality) by directly substituting numbers from the ODZ.

Example 1. Solve the equation

Solution. The ODZ of this equation consists of all x that simultaneously satisfy the conditions 3-x0 and x-3>0, that is, the ODZ is an empty set. This completes the solution of the equation, since it has been established that not a single number can be a solution, that is, that the equation has no roots.

Answer: there are no solutions.

Example 2. Solve the equation

(1)

Solution. The ODZ of this equation consists of all x that simultaneously satisfy the conditions, that is, the ODZ is Substituting these values ​​of x into equation (1), we find that its left and right sides are equal to 0, which means that all https://pandia.ru/ text/78/500/images/image065_2.gif" width="93 height=21" height="21">

Example 3. Solve inequality

Solution. The ODZ of inequality (2) consists of all x that simultaneously satisfy the conditions that is, the ODZ consists of two numbers and . Substituting into inequality (2), we find that its left side is equal to 0, the right side is equal to https://pandia.ru/text/78/500/images/image070_1.gif" width="53" height="23">. gif" width="117 height=41" height="41">.

Answer: x=1.

Example 4. Solve inequality

(3)

Solution. The ODZ of inequality (3) is all x satisfying the condition 0<х1. Ясно, что х=1 не является решением неравенства (3). Для х из промежутка 0

Answer: 0

Example 5. Solve inequality

Solution..gif" width="73" height="19"> and .

For x from the interval https://pandia.ru/text/78/500/images/image082_1.gif" width="72" height="24 src=">.gif" width="141 height=24" height= "24"> on this interval, and therefore inequality (4) has no solutions on this interval.

Let x belong to the interval, then https://pandia.ru/text/78/500/images/image087_1.gif" width="141 height=24" height="24"> for such x, and, therefore, on In this interval, inequality (4) also has no solutions.

So, inequality (4) has no solutions.

Answer: there are no solutions.

Notes.

When solving equations, it is not necessary to find the ODZ. Sometimes it’s easier to move on to the investigation and check the roots found. When solving inequalities, sometimes it is possible not to find the ODZ, but to solve the inequality by moving to an equivalent system of inequalities, in which either one of the inequalities has no solutions, or knowledge of its solution helps solve the system of inequalities.

Example 6. Solve inequality

Solution. Finding the ODZ of inequality is not an easy task, so we will do it differently. Inequality (5) is equivalent to the system of inequalities

(6)

The third inequality of this system is equivalent to an inequality that has no solutions. Consequently, the system of inequalities (6) has no solutions, which means inequality (5) has no solutions.

Answer: there are no solutions.

Example 7. Solve inequality

. (7)

Solution. Finding the ODZ of inequality (7) is a difficult task. Therefore, let's do things differently. Inequality (7) is equivalent to the system of inequalities

(8)

The third inequality of this system has solutions to all x from the interval -1

2.2. Exploiting limited functionality.

When solving equations and inequalities, the property of a function being bounded below or above on a certain set often plays a decisive role.

For example, if for all x from some set M the following inequalities are true: f(x)>A and g(x)

Note that the role of the number A is often played by zero; in this case, they say that the sign of the functions f(x) and g(x) on the set M is preserved.

Example 1. Solve the equation

Solution..gif" width="191" height="24 src="> Since for any value of x the left side of the equation does not exceed one, and the right side is always not less than two, then this equation has no solutions.

Answer: no solutions.

Example 2. Solve the equation

(9)

Solution. It is obvious that x=0, x=1, x=-1 are solutions to equation (9)..gif" width="36" height="19">, since if is its solution, then (-) is also his decision.

Let us divide the set x>0, , into two intervals (0;1) and (1;+∞).

Let's rewrite equation (9) in the form https://pandia.ru/text/78/500/images/image103_1.gif" width="104" height="28">.gif" width="99" height="25 src=">only positive ones. Consequently, on this interval, equation (9) has no solutions.

Let x belong to the interval (1;+∞). For each of these values ​​x the function takes positive values, the function https://pandia.ru/text/78/500/images/image105_1.gif" width="99" height="25 src="> is non-positive. Therefore, on this interval, equation (9) is not has solutions.

If x>2, then , and this means that on the interval (2;+∞) equation (9) also has no solutions.

So, x=0, x=1 and x=-1 and only these are solutions to the original equation.

Answer:

Example 3. Solve inequality

Solution. The ODZ of inequality (10) is all real x, except x=-1. Let us divide the ODZ into three sets: -∞<х<-1, -1

Let -∞<х<-1..gif" width="93" height="24 src=">. Consequently, all these x are solutions to inequality (10).

Let -1 , A . Consequently, none of these x is a solution to inequality (10).

Let 0 , A . Consequently, all these x are solutions to inequality (10).

Answer: -∞<х<-1; 0

Example 4. Solve the equation

(11)

Solution. Let's denote via f(x). From the definition of absolute value it follows that f(x)= at , https://pandia.ru/text/78/500/images/image120_1.gif" width="84" height="41 src=">.gif" width="43" height="41 src=">. Therefore, if , then equation (11) can be rewritten in the form , that is, in the form ..gif" width="53" height="41"> satisfy only . If , then equation (11) can be rewritten as https://pandia.ru/text/78/500/images/image128_0.gif" width="73 height=41" height="41">. This equation has solutions . Of these x values, only .

Consider x from the interval . On this interval, equation (11) can be rewritten in the form , that is, in the form

It is clear that x = 0 is a solution to equation (12), and therefore to the original equation..gif" width="39" height="19"> equation (12) is equivalent to the equation

For any value , the function takes only positive values, therefore equation (12) has no solutions on the set .

Answer: x=0, ; https://pandia.ru/text/78/500/images/image139_0.gif" width="211" height="41">. (13)

Solution. Let there be a solution to equation (13), then the following equality holds: (14)

and inequalities https://pandia.ru/text/78/500/images/image142_1.gif" width="68" height="27 src=">. From the validity of the inequalities we obtain that the left side of equality (14) has the the same sign as , that is, the same sign as , and the right side is the same sign as , but since and satisfy equality (14), they have the same signs.

Let us rewrite equality (14) in the form

https://pandia.ru/text/78/500/images/image147_0.gif" width="284" height="24">

Let us rewrite equality (15) in the form

Since they have the same signs, then ..gif" width="95" height="24">. (17)

It is obvious that any solution to equation (17) is a solution to equation (13). Therefore, equation (13) is equivalent to equation (17). The solutions to equation (17) are , they and only they are solutions to equation (13).

Answer:

Comment. Just as in Example 5, it can be proven that Eq.

where n, m are any natural numbers, is equivalent to the equation, and then solve this simpler equation.

2. 3. Using the monotonicity of the function.

Solving equations and inequalities using the monotonicity property is based on the following statements.

Let f(x) be a continuous and strictly monotone function on the interval L, then the equation f(x)=C, where C is a given constant, can have at most one solution on the interval L. Let f(x) and g(x) are continuous functions on the interval L, f(x) strictly increases, and g(x) strictly decreases on this interval, then the equation f(x)=g(x) can have at most one solution on the interval L.

Note that the interval L can be an infinite interval (-∞; +∞), semi-infinite intervals (a; +∞), (-∞; a), [a; +∞), (-∞; a], segments, intervals and half-intervals.

Example 1. Solve the equation

(18)

Solution. Obviously, x0 cannot be a solution to equation (18), since then . For x>0 function continuous and strictly increasing, as the product of two continuous positive strictly increasing functions f=x and https://pandia.ru/text/78/500/images/image157_0.gif" width="119" height="34" > takes each of its values ​​at exactly one point.It is easy to see that x=1 is a solution to equation (18), therefore, this is its only solution.

Answer: x=1.

Example 2. Solve inequality

. (19)

Solution. Each of the functions , , is continuous and strictly increasing on the entire axis. This means that the original function is the same . It is easy to see that for x=0 the function takes the value 3. Due to the continuity and strict monotonicity of this function for x>0 we have , at x<0 имеем . Consequently, solutions to inequality (19) are all x<0.

Answer: -∞

Example 3. Solve the equation

(20)

Solution. The range of permissible values ​​of equation (20) is the interval . On the range of valid values ​​of the function And are continuous and strictly decreasing, therefore, the function is continuous and decreasing. Therefore, the function h(x) takes each value only at one point. Since h(2)=2, then x=2 is the only root of the original equation.

Answer: x=2.

Example 4. Solve inequality

The solution... gif" width="95" height="25 src="> are presented in Figure 7. From the figure it follows that for all x from the ODZ inequality (26) is valid.

Let's prove it. For each we have , and for each such x we ​​have that https://pandia.ru/text/78/500/images/image211_1.gif" width="63 height=23" height="23"> we have . Consequently, the solutions to inequality (26) will be all x from the interval [-1;1].

Example 2. Solve the equation

. (27)

Solution..gif" width="123" height="24"> and are presented in Figure 8. Let's draw a straight line y=2. It follows from the figure that the graph of the function f(x) lies no lower than this line, and the graph of the function g(x) no higher. Moreover, these graphs touch the straight line y=2 at different points. Therefore, the equation has no solutions. Let's prove it. For each we have , a . In this case, f(x)=2 only for x=-1, and g(x)=2 only for x=0. This means that equation (27) has no solutions.

Answer: there are no solutions.

Example 3. Solve the equation

. (28)

Solution..gif" width="95" height="25 src="> are presented in Figure 9. It is easy to check that point (-1; -2) is the intersection point of the graphs of the functions f(x) and g(x), that is, x=-1 is the solution to equation (28). Let's draw a straight line y=x-1. From the figure it follows that it is located between the graphs of the functions y=f(x) and y=g(x). This observation helps to prove that equation (28) has no other solutions.

To do this, we prove that x from the interval (-1; +∞) the inequalities and , and for x from the interval (-∞; -1) the inequalities https://pandia.ru/text/78/500/images/image229_1 .gif" width="89" height="21 src=">. Obviously, the inequality is valid for x>-1, and the inequality https://pandia.ru/text/78/500/images/image228_1.gif" width="93" height="24">..gif" width="145" height="25">. The solutions to this inequality are all x<-1. Точно так же показывается, что решениями неравенства являются все х>-1.

Consequently, the required statement is proven, and equation (28) has a single root x=-1.

Answer: x=-1.

Example 4. Solve inequality

. (29)

Solution..gif" width="39" height="19 src=">, that is, the ODZ consists of three spaces, , https://pandia.ru/text/78/500/images/image234_1.gif" width= "52" height="41">, is equivalent to inequality

, (30)

and in the region x>0 it is equivalent to the inequality

. (31)

Function graph sketches and are shown in Figure 10..gif" width="56" height="45"> and .

Therefore, inequality (31) has no solutions, and inequality (30) will have solutions to all x from the interval .

Let's prove it.

A) Let . Inequality (29) is equivalent to inequality (30) on this interval. It is easy to see that for each x from this interval the inequalities are valid

,

.

Consequently, inequality (30), and along with it the original inequality (29), have no solutions on the interval .

B) Let . Then inequality (29) is also equivalent to inequality (30). For each x from this interval

,

Consequently, any such x is a solution to inequality (30), and therefore to the original inequality (29).

C) Let x>0. On this set, the original inequality is equivalent to inequality (31). Obviously, for any x from this set the following inequalities are true:

,

This implies:

1) inequality (31) has no solutions on the set where , that is, inequality (31) has no solutions on the set ;

2) inequality (31) has no solutions on the set where https://pandia.ru/text/78/500/images/image253_1.gif" width="60" height="19">. It remains to find solutions to inequality (31 ) belonging to interval 1