Solving Odz inequalities. ODZ. Area of Acceptable Values. Respecting your privacy at the company level
When solving various problems, we very often have to carry out identical transformations of expressions. But it happens that some kind of transformation is acceptable in some cases, but not in others. Significant assistance in terms of monitoring the admissibility of ongoing transformations is provided by ODZ. Let's look at this in more detail.
The essence of the approach is as follows: the ODZ of variables for the original expression is compared with the ODZ of variables for the expression obtained as a result of identical transformations, and based on the comparison results, appropriate conclusions are drawn.
In general, identity transformations can
- do not influence DL;
- lead to the expansion of ODZ;
- lead to a narrowing of ODZ.
Let's illustrate each case with an example.
Consider the expression x 2 +x+3·x, the ODZ of the variable x for this expression is the set R. Now let's do the following identical transformation with this expression - we present similar terms, as a result it will take the form x 2 +4·x. Obviously, the variable x of this expression is also a set R. Thus, the transformation carried out did not change the DZ.
Let's move on. Let's take the expression x+3/x−3/x. In this case, the ODZ is determined by the condition x≠0, which corresponds to the set (−∞, 0)∪(0, +∞) . This expression also contains similar terms, after reducing which we arrive at the expression x, for which the ODZ is R. What we see: as a result of the transformation, the ODZ was expanded (the number zero was added to the ODZ of the variable x for the original expression).
It remains to consider an example of narrowing the range of acceptable values after transformations. Let's take the expression 
. The ODZ of the variable x is determined by the inequality (x−1)·(x−3)≥0, for its solution it is suitable, for example, as a result we have (−∞, 1]∪∪; edited by S. A. Telyakovsky. - 17- ed. - M.: Education, 2008. - 240 pp.: ill. - ISBN 978-5-09-019315-3.
If the ODZ of an equation consists of a finite number of values, it is enough to substitute each value into the equation to check whether this value is a root.
Examples of applying the finite to solving equations.
The sign of an even root must have a non-negative number, so
The first inequality is quadratic, let's solve it. Second - .
The solution to the system is the intersection of the solutions to both inequalities:
ODZ consists of a single value: (3).
It remains to check whether 3 is the root of the equation:
We got the correct equality, therefore x=3 is the root of this equation.
The square root sign must have a non-negative number. Hence the ODZ
The first two inequalities are quadratic. We solve them using the interval method. The third is linear. We mark the solution to each inequality on the number line and find the intersection of the solutions:
ODZ consists of two values: (2; 3).
Let's check.
Thus, this equation has a single root x=3.
The range of permissible arcsine values is the closed interval from -1 to 1. The base of a power with a non-integer positive exponent must be a non-negative number. ODZ:
Thus, the range of acceptable values of the equation consists of one value: (1). It remains to check whether x=1 is a root of this equation.
Answer: 1.
If the ODZ of the equation consists of one or more numbers, this method can help you cope with the task easily and quickly.
Like other methods for solving equations based on the properties of functions, the use of a finite number of values often allows one to solve quite complex non-standard tasks. And although it does not appear often in the school algebra course, it is useful to remember it and be able to apply it.
Category: |Scientific adviser:
1. Introduction 3
2. Historical sketch 4
3. “Place” of ODZ when solving equations and inequalities 5-6
4. Features and dangers of ODZ 7
5. ODZ – there is a solution 8-9
6. Finding ODZ is extra work.
Equivalence of transitions 10-13
7. ODZ in the Unified State Exam 14-15
8. Conclusion 16
9. Literature 17
1. Introduction
Equations and inequalities in which it is necessary to find the range of acceptable values have not found a place in the algebra course of systematic presentation, which is perhaps why my peers often make mistakes when solving such examples, spending a lot of time solving them, while forgetting about the range of acceptable values. This determined problem of this work.
In this work it is intended to investigate the phenomenon of the existence of a region of acceptable values when solving equations and inequalities of various types; analyze this situation, draw logically correct conclusions in examples where you need to take into account the range of acceptable values.
Tasks:
- Based on existing experience and theoretical basis, collect basic information about the range of permissible values and its use in school practice; Analyze solutions to various types of equations and inequalities (fractional-rational, irrational, logarithmic, containing inverse trigonometric functions); Check the results previously obtained when solving various equations and inequalities, make sure of the reliability of the methods and methods for solving them; Determine the “place” of the range of acceptable values when solving equations and inequalities; Apply the research materials obtained in a situation that differs from the standard one, and use them in preparation for the Unified State Exam.
When solving these problems, the following were used research methods: analysis, statistical analysis, deduction, classification, forecasting.
The study began with a repetition of well-known functions studied in the school curriculum. The scope of many of them is limited.
The range of acceptable values occurs when solving: fractional rational equations and inequalities; irrational equations and inequalities; logarithmic equations and inequalities; equations and inequalities containing inverse trigonometric functions.
Having solved many examples from various sources (USE textbooks, textbooks, reference books), we identified the solution of examples according to the following principles:
· you can solve the example and take into account the ODZ (the most common method)
· it is possible to solve the example without taking into account the ODZ
· it is only possible to come to the right decision by taking into account the ODZ.
An analysis of the Unified State Examination results over the past years has been studied. Many mistakes were made in examples in which it is necessary to take into account DL. Practical significance The work lies in the fact that its content, assessments and conclusions can be used in teaching mathematics at school, in preparation for the final certification of schoolchildren in grades 9 and 11.
2. Historical sketch
Like other concepts of mathematics, the concept of a function did not develop immediately, but went through a long path of development. In the work of P. Fermat “Introduction and study of plane and solid places” (1636, published 1679) it is said: “Whenever there are two unknown quantities in the final equation, there is a place.” Essentially, here we are talking about functional dependence and its graphical representation (“place” in Fermat means a line). The study of lines according to their equations in R. Descartes' "Geometry" (1637) also indicates a clear understanding of the mutual dependence of two variables. I. Barrow (Lectures on Geometry, 1670) establishes in geometric form the mutual inverse nature of the actions of differentiation and integration (of course, without using these terms themselves). This already indicates a completely clear mastery of the concept of function. We also find this concept in geometric and mechanical form in I. Newton. However, the term “function” first appears only in 1692 with G. Leibniz and, moreover, not quite in its modern understanding. G. Leibniz calls various segments associated with a curve (for example, the abscissa of its points) a function. In the first printed course, “Analysis of infinitesimals for the knowledge of curved lines” by L'Hopital (1696), the term “function” is not used.
The first definition of a function in a sense close to the modern one is found in I. Bernoulli (1718): “A function is a quantity composed of a variable and a constant.” This not entirely clear definition is based on the idea of specifying a function by an analytical formula. The same idea appears in the definition of L. Euler, given by him in “Introduction to the Analysis of Infinites” (1748): “The function of a variable quantity is an analytical expression composed in some way from this variable quantity and numbers or constant quantities.” However, L. Euler is no longer alien to the modern understanding of function, which does not connect the concept of a function with any of its analytical expressions. His “Differential Calculus” (1755) says: “When certain quantities depend on others in such a way that when the latter change they themselves are subject to change, then the former are called functions of the latter.”
Since the beginning of the 19th century, the concept of a function has been increasingly defined without mentioning its analytical representation. In the “Treatise on Differential and Integral Calculus” (1797-1802) S. Lacroix says: “Every quantity whose value depends on one or many other quantities is called a function of these latter.” In the “Analytical Theory of Heat” by J. Fourier (1822) there is a phrase: “Function f(x) denotes a completely arbitrary function, that is, a sequence of given values, whether or not subject to a general law and corresponding to all values x contained between 0 and some value x" The definition of N. I. Lobachevsky is close to the modern one: “...The general concept of a function requires that a function from x name the number that is given for each x and together with x gradually changes. The value of the function can be given either by an analytical expression, or by a condition that provides a means of testing all the numbers and choosing one of them, or, finally, the dependence can exist and remain unknown. It is also said there a little lower: “The broad view of the theory allows for the existence of dependence only in the sense that numbers one with another in connection are understood as if given together.” Thus, the modern definition of a function, free of reference to the analytical task, usually attributed to P. Dirichlet (1837), was repeatedly proposed before him:
y has a function of the variable x (on the segment https://pandia.ru/text/78/093/images/image002_83.gif" width="95" height="27 src=">. By squaring both sides of the equation, we let's get rid of the irrationality. But let's pay attention to the fact that squaring, generally speaking, is not an equivalent transformation, and when squaring we can get extra roots. If the roots are whole, then it is easy to check. But in some cases it is inconvenient to check Then use the reduction of this equation to an equivalent system:
.
In this case, there is no need to find the ODZ: from the first equation it follows that the obtained values of x satisfy the following inequality: https://pandia.ru/text/78/093/images/image005_34.gif" width="107" height="27 src="> is the system:
Since they enter into the equation equally, then instead of inequality, you can include inequality https://pandia.ru/text/78/093/images/image010_15.gif" width="220" height="49"> 
https://pandia.ru/text/78/093/images/image015_10.gif" width="239" height="51"> 
3. Solving logarithmic equations and inequalities.
3.1. Scheme for solving a logarithmic equation
But it is enough to check only one condition of the ODZ.
3.2..gif" width="115" height="48 src=">.gif" width="115" height="48 src=">
4. Trigonometric equations of the form are equivalent to the system (instead of inequality, you can include inequality in the system https://pandia.ru/text/78/093/images/image025_2.gif" width="377" height="23"> are equivalent to the equation
4. Features and dangers of the range of permissible values
In mathematics lessons, we are required to find the DL in each example. At the same time, according to the mathematical essence of the matter, finding the ODZ is not at all mandatory, often not necessary, and sometimes impossible - and all this without any damage to the solution of the example. On the other hand, it often happens that after solving an example, schoolchildren forget to take into account the DL, write it down as the final answer, and take into account only some conditions. This circumstance is well known, but the “war” continues every year and, it seems, will continue for a long time.
Consider, for example, the following inequality:
Here, the ODZ is sought and the inequality is solved. However, when solving this inequality, schoolchildren sometimes believe that it is quite possible to do without searching for ODZ, or more precisely, it is possible to do without the condition
In fact, to obtain the correct answer it is necessary to take into account both the inequality , and .
But, for example, the solution to the equation: https://pandia.ru/text/78/093/images/image033_3.gif" width="79 height=75" height="75">
which is equivalent to working with ODZ. However, in this example, such work is unnecessary - it is enough to check the fulfillment of only two of these inequalities, and any two.
Recall that any equation (inequality) can be reduced to the form . ODZ is simply the domain of definition of the function on the left side. The fact that this area must be monitored follows from the definition of the root as a number from the domain of definition of a given function, thereby from the ODZ. Here is a funny example on this topic..gif" width="20" height="21 src="> has a domain of definition of a set of positive numbers (this, of course, is an agreement to consider a function with, but reasonable), and then -1 is not is the root.
5. Range of acceptable values – there is a solution
And finally, in a lot of examples, finding ODZ allows you to get an answer without cumbersome calculations, or even orally.
1. OD3 is an empty set, which means that the original example has no solutions.
1)
2) 3) 
2. B ODZ one or more numbers are found, and a simple substitution quickly determines the roots.
1)
, x=3
2)
Here in the ODZ there is only a number 1
, and after substitution it is clear that it is not a root.
3) There are two numbers in the ODZ: 2 And 3 , and both are suitable.
4) > There are two numbers in the ODZ 0 And 1 , and only fits 1 .
ODZ can be used effectively in combination with analysis of the expression itself.
5) 
< ОДЗ: Но в правой части неравенства могут быть только положительные числа, поэтому оставляем x=2. Then we substitute into the inequality 2
.
6) 
From the ODZ it follows that, where we have ..gif" width="143" height="24"> From the ODZ we have: . But then and . Since, there are no solutions.
From the ODZ we have:..gif" width="53" height="24 src=">.gif" width="156" height="24"> ODZ: . Since then
On the other side,. Equality is only possible when each side of the equation is equal 0 , i.e. when x=1. After substituting this value X We are convinced that there are no solutions.
ODZ:. Consider the equation on the interval [-1; 0).
It fulfills the following inequalities https://pandia.ru/text/78/093/images/image072_0.gif" width="68" height="24 src=">.gif" width="123" height="24 src="> and there are no solutions. With the function and https://pandia.ru/text/78/093/images/image077_0.gif" width="179" height="25">. ODZ: x>2. Wherein . This means that the initial equality is impossible and there are no solutions.
Now let’s give an example that was suggested by the teacher in an algebra lesson. We weren’t able to solve it right away, but when we found the ODZ, everything became clear.
Find the integer root of the equation https://pandia.ru/text/78/093/images/image080_0.gif" width="124" height="77"> 
An integer solution is only possible if x=3 And x=5. By checking we find that the root x=3 doesn't fit, so the answer is: x=5.
6. Finding the range of acceptable values is extra work. Equivalence of transitions.
You can give examples where the situation is clear even without finding DZ.
1. 
Equality is impossible, because when subtracting a larger expression from a smaller one, the result must be a negative number.
2. 
.
The sum of two non-negative functions cannot be negative.
I will also give examples where finding ODZ is difficult, and sometimes simply impossible.
And finally, searches for ODZ are very often just extra work, which you can do without, thereby proving your understanding of what is happening. There are a huge number of examples that can be given here, so we will select only the most typical ones. The main solution method in this case is equivalent transformations when moving from one equation (inequality, system) to another.
1.
. ODZ is not needed, because, having found those values X, at which x2=1, we can't get x=0.
2. . ODZ is not needed, because we find out when the radical expression is equal to a positive number.
3. . ODZ is not needed for the same reasons as in the previous example.
4. 
ODZ is not needed, because the radical expression is equal to the square of some function, and therefore cannot be negative.
5. 
6. . ODZ is not needed, since the expression is always positive.
7. 
To solve, only one constraint for the radical expression is sufficient. In fact, from the written mixed system it follows that the other radical expression is non-negative.
8. DZ is not needed for the same reasons as in the previous example.
9. 
ODZ is not needed, since it is enough for two of the three expressions under the logarithm signs to be positive to ensure the positivity of the third.
10. 
.gif" width="357" height="51"> ODZ is not needed for the same reasons as in the previous example.
It is worth noting, however, that when solving using the method of equivalent transformations, knowledge of the ODZ (and properties of functions) helps.
Here are some examples.
1. . OD3, which implies that the expression on the right side is positive, and it is possible to write an equation equivalent to this one in this form. The obtained result must be checked against the ODZ.
2. 
ODZ: . But then, and when solving this inequality, it is not necessary to consider the case when the right-hand side is less than 0.
3. . It follows from the ODZ that , and therefore the case when , is excluded.
In general, the effectiveness of the method of equivalent transformations seems to be clear. With their help, we get to the answer without searching for DZ. Does this mean that there is some universal method and all that remains is to learn how to use it? But it is not so. There are several reasons for this. There are quite a lot of theorems about equivalent transformations, they are not easy to remember, and confident mastery of them is not an easy matter. Often, using equivalent transformations, you begin to put this sign on any transitions from one equation to another, both truly equivalent and not so. These theorems are quickly forgotten.
Another difficulty is that when writing equivalence, you can forget to write down all the conditions that guarantee it, but this may not affect the answer in any way. Here are two such examples:
1. The transition in general looks like this:
In this example, the expression under the sign of the logarithm on the right is always positive. Therefore, in relation to this example, that part of the equivalence conditions that is written as a set does not add anything. But having made such a decision, you can simply forget about this totality.
There are two possible cases: 0<<1 и >1.
This means that the original inequality is equivalent to the following set of systems of inequalities:
The first system has no solutions, but from the second we obtain: x<-1 – решение неравенства.
Understanding the conditions of equivalence requires knowledge of some subtleties. For example, why are the following equations equivalent:
Or 
And finally, perhaps most importantly. The fact is that equivalence guarantees the correctness of the answer if some transformations of the equation itself are made, but is not used for transformations in only one of the parts. Abbreviations and the use of different formulas in one of the parts are not covered by the equivalence theorems. Some examples of this type were given in the work. Let's look at some more examples.
1. This decision is natural. On the left side, according to the property of the logarithmic function, we move on to the expression. As a result, we obtain the equation. It is equivalent to such a system
Having solved this system, we get the result (-2 and 2), which, however, is not an answer, since the number -2 is not included in the ODZ. So, do we need to install ODS? Of course not. But since we used a certain property of the logarithmic function in the solution, then we are obliged to provide the conditions under which it is satisfied. Such a condition is the positivity of expressions under the logarithm sign..gif" width="65" height="48">.
2. 
..gif" width="143" height="27">.gif" width="147" height="24">add a condition, and you can immediately see that only the number https://pandia.ru/ meets this condition text/78/093/images/image129.gif" width="117" height="27">) were demonstrated by 52% of those taking the test. One of the reasons for such low rates is the fact that many graduates did not select the roots obtained from the equation after squaring it.
3) Consider, for example, the solution to one of the problems C1: “Find all values of x for which the points of the graph of the function 
lie above the corresponding points on the graph of the function. The task boils down to solving a fractional inequality containing a logarithmic expression. We know the methods for solving such inequalities. The most common of them is the interval method. However, when using it, test takers make various mistakes. Let's look at the most common mistakes using inequality as an example:
1. Graduates correctly find the DL by solving the system of inequalities:
where x 
. Next, multiplying both sides of the inequality by a common denominator, we obtain the inequality: log(23 - 10 x
2..gif" width="124" height="29">. Next they get x– 10 +
; . By solving this equation and taking into account the condition, graduates conclude that the equation has no solutions.
3. Test takers correctly transform the equation to the form
and consider two cases: x 10 and x < 10. Они отмечают, что в первом случае решений нет, а во втором – корнями являются числа –1 и . При этом выпускники не учитывают условие x < 10.
8. Conclusion
In this work, we tried to explore the phenomenon of the existence of a range of acceptable values when solving equations and inequalities of various types, analyzed this situation, and made logically correct conclusions in examples where it is necessary to take into account the range of acceptable values. For me, the topic “Region of Permissible Values” seemed very complex and incomprehensible, and in school textbooks this topic is not given due place, it is practically not covered, although the Unified State Examination tasks contain tasks on solving equations and inequalities in which it is necessary to find the range of permissible values . In the process of work, we were faced with the fact that the literature on this topic is not enough for a complete and systematic study. We think that this topic requires close attention of mathematicians and methodologists.
Having solved many examples from various sources, we can draw some conclusions: there is no universal method for solving equations and inequalities. Every time, if you want to understand what you are doing and not act mechanically, you think: what solution method should you choose, in particular, should you look for the range of acceptable values or should you not? We believe that the experience gained will help solve this dilemma. Students will stop making mistakes by learning to correctly use the range of acceptable values. Whether we will be able to do this, time will tell, or rather the upcoming Unified State Exam 2010.
We hope that the presented work will be interesting and useful to teachers and students, and that educational disabilities will cease to be “some kind of bad ODZ” for schoolchildren.
9. Literature
1., etc. “Algebra and the beginnings of analysis 10-11” problem book and textbook, M.: “Prosveshchenie”, 2002.
2. “Handbook of elementary mathematics.” M.: “Science”, 1966.
3. Newspaper “Mathematics” No. 46,
4. Newspaper "Mathematics" No.
5. Newspaper "Mathematics" No.
6. “History of mathematics in school, grades VII-VIII.” M.: “Enlightenment”, 1982.
7. etc. “The most complete edition of options for real Unified State Examination tasks: 2009/FIPI” - M.: “Astrel”, 2009.
8. and others “Unified State Examination. Mathematics. Universal materials for preparing students/FIPI" - M.: "Intellect Center", 2009.
9. and others. “Algebra and the beginnings of analysis 10-11.” M.: “Enlightenment”, 2007.
10. , “Workshop on solving problems in school mathematics (workshop in algebra).” M.: Education, 1976.
11. “25,000 math lessons.” M.: “Enlightenment”, 1993.
12. “We are preparing for the Olympiads in mathematics.” M.: “Exam”, 2006.
13. “Encyclopedia for children “MATHEMATICS”” volume 11, M.: Avanta +; 2002.
14. Materials from the sites http://www. *****, http://www. *****.
Internet portal Wikipedia http://ru. wikipedia. org/wiki/Numeric_function (Accessed 03/05/2010).
, “Workshop on solving problems in school mathematics (workshop in algebra).” M.: Education, 1976, p.64.
Question from a student to Answers@***** http://otvet. *****/question/8166619/ (Date viewed 03/22/2010)
Methodological letter “On the use of the results of the 2008 Unified State Exam in teaching mathematics in educational institutions of secondary (complete) general education” http://www. ***** (Date viewed 12/17/2009)
Congratulations, dear readers!
We've finally reached solving trigonometric equations. Now we will solve several equations that are similar to the Unified State Examination tasks. Of course, in the real exam, the tasks will be a little more difficult, but the essence will remain the same.
First, let's look at an easy equation (we've already solved similar ones in previous lessons, but repeating them is always useful).
$$(2\cos x + 1) (2\sin x - \sqrt(3)) = 0.$$
I think explanations of how to decide are unnecessary.
$$2\cos x + 1 = 0 \text( or ) 2\sin x - \sqrt(3) =0,$$
$$\cos x = -\frac(1)(2) \text( or ) \sin x = \frac(\sqrt(3))(2),$$
The horizontal dotted line marks solution for equation with sine, vertical - with cosine.
Thus, the final solution can be written, for example, like this:
$$\left[ \begin(array)(l)x= \pm \frac(2\pi)(3),\\x = \frac(\pi)(3)+2\pi k. \end(array)\right.$$
Trigonometric equation with ODZ
$$(1+\cos x)\left(\frac(1)(\sin x) - 1\right) = 0.$$
An important difference in this example is that a sine appears in the denominator. Although we have solved similar equations a little in previous lessons, it is worth dwelling on the ODZ in more detail.
ODZ
`\sin x \neq 0 \Rightarrow x \neq \pi k`. When we mark the solution on the circle, we will mark this series of roots with specially pierced (open) points to show that `x` cannot take such values.
Solution
Let us reduce to a common denominator, and then alternately equate both brackets to zero.
$$(1+\cos x)\left(\frac(1-\sin x)(\sin x)\right) = 0,$$
$$1+\cos x = 0 \text( or ) \frac(1-\sin x)(\sin x) = 0,$$
$$\cos x = -1 \text( or ) \sin x=1.$$
I hope solving these equations will not cause any difficulties.
The series of roots - solutions to the equation - are shown below with red dots. ODZ is marked in blue in the figure.
Thus, we understand that the solution to the equation `\cos x = -1` does not satisfy the ODZ.
The answer will only be a series of roots `x = \frac(\pi)(2) + 2\pi k`.
Solving a quadratic trigonometric equation
The next point in our program is solving a quadratic equation. There is nothing complicated about it. The main thing is to see the quadratic equation and make the replacement as shown below.
$$3\sin^2 x + \sin x =2,$$
$$3\sin^2 x + \sin x -2=0.$$
Let `t= \sin x`, then we get:
$$3t^2 + t-2=0.$$
$$t_1 = \frac(2)(3), t_2 = -1.$$
Let's do the reverse replacement.
$$\sin x = \frac(2)(3) \text( or ) \sin x = -1.$$
$$\left[\begin(array)(l)x = \arcsin \frac(2)(3) + 2\pi k, \\ x = \pi - \arcsin \frac(2)(3) + 2 \pi k, \\ x = -\frac(\pi)(2) + 2\pi k. \end(array) \right.$$
Solving a quadratic equation with tangent
Let's solve the following equation:
$$\newcommand(\tg)(\mathop(\mathrm(tg)))(\tg)^2 2x - 6\tg 2x +5 =0, $$
Please note that the tangent argument is `2x` and to get the final answer you will need to divide by `2`. Let `t=\tg 2x`, then
$$t^2 - 6t +5 =0, $$
$$t_1 = 5, t_2 = 1.$$
Reverse replacement.
$$\tg x = 5,\tg x = 1.$$
$$\left[\begin(array)(l)2x = \arctan(5)+\pi k, \\ 2x = \frac(\pi)(4) + \pi k. \end(array) \right.$$
Now let's divide both series by two to find out what `x` actually equals.
$$\left[\begin(array)(l)x = \frac(1)(2)\arctan(5)+\frac(\pi k)(2), \\ 2x = \frac(\pi) (8) + \frac(\pi k)(2). \end(array) \right.$$
So we got the answer.
Last equation (product of tangent and sine)
$$\tg x \cdot \sin 2x = 0.$$
ODZ
Since the tangent is a fraction whose denominator is the cosine, then in the ODZ we obtain that `\cos x \neq 0 \Rightarrow x \neq \frac(\pi)(2)+\pi k.`
Solution
$$\tg x =0 \text( or ) \sin 2x = 0.$$
These equations are easy to solve. We get:
$$x = \pi k \text( or ) 2x = \pi k,$$
$$x = \pi k \text( or ) x = \frac(\pi k)(2).$$
Now the most interesting thing: since we had ODZ, we need to perform a selection of roots. Let us mark the resulting series of roots on a circle. (How to do this is shown in detail in the attached video.)
ODZ is marked in blue, solutions are in red. It can be seen that the answer will be `x = \pi k`.
This concludes the fifth lesson. Be sure to practice solving equations. It’s one thing to know the progress of a solution in general terms, it’s another thing to get your bearings when solving a specific problem. Gradually practice each element of solving the problem. Now the main thing is to learn how to work competently with the trigonometric circle, find solutions with its help, see the ODZ and correctly make substitutions for quadratic equations.
Tasks for training
Solve the equations:
- `2 \cos^2 \frac(x)(2) + \sqrt(3) \cos \frac(x)(2) = 0`,
- `3 (\tg)^2 2x + 2\tg 2x -1= 0`,
- `2\cos^2 3x - 5\cos 3x -3 =0`,
- `\sin^2 4x + \sin x - \cos^2x =0` (apply the basic trigonometric identity),
- `4\sin^2 \left(x-\frac(\pi)(3) \right) - 3 =0`.
That's enough. If you have any questions, just ask! Leave a like if my work was useful :)
How to look for this same ODZ? We carefully examine the example and look for dangerous places. Places where prohibited actions are possible. There are very few such forbidden actions in mathematics.
More lessons on the site
ARV (Acceptable Value Area)
The range of acceptable values of an equation is the set of values of x for which the right and left sides of the equation make sense.
These are the values of x that can be in principle. Let's say in the equation = 1 we don't yet know what x is equal to. We haven't solved the equation yet. But we already know for sure that x cannot be equal to zero under any circumstances! You can't divide by zero! Any other number - integer, fractional, negative - please, but zero - never! Otherwise the original expression becomes nonsense. This means that the ODZ in this example is: x – anything other than zero. Got it?
How to find, how to record, how to work with it?
Very simple. Write ODZ next to the example. Under these well-known letters, looking at the original equation, we write down the values of x, which are allowed for the original example. Or vice versa: find forbidden values of x, in which the original example loses all meaning, and exclude them.
But not everyone remembers them either. I’ll remind you of them now, and I advise you to remember them.
The expression under the root sign of even multiplicity must be greater than or equal to zero.
The expression in the denominator of a fraction cannot be equal to zero.
- There are two functions that contain a "hidden" fraction:
There are also prohibitions in logarithmic equations - we will look at these in the relevant topics. All. When we have found dangerous places, we calculate x, which will lead to nonsense.
To find the range of acceptable expression values, you need to examine whether there areexpression equation which I listed above. And as you discover expressions, write down the restrictions they set, moving “outside” “inside.” And we exclude them.
Important! To find ODZ we do not solve an example! We solve pieces of the example to find forbidden X's. This looks difficult in explanation, but in practice it is very easy.
I specifically didn’t say anything about DD in previous lessons. So as not to scare you away... In the examples considered, DL did not affect the answers in any way. After all, in our listed prohibitions there is no demonstrative function. It happens. But in tasks for EXTERNAL INDEPENDENT TESTING, DL, as a rule, affects the answer! It should be written not for inspectors, but for yourself. do not write if it is obvious that x is any number. As, for example, in linear equations.
In a lot of examples, finding ODZ allows you to get an answer without cumbersome calculations. Or even orally. In some equations, it represents an empty set. This means that the original equation has no solutions. Or there is one or more numbers in there, and a simple substitution quickly determines the roots.
What's not to like? That's right - a fraction. I don't like it either, so I suggest getting rid of it. This can be done in different ways. In order to get rid of the denominator, I will multiply both sides of the equation by the common denominator x-4.
Loading...
