Equation ctg x 1. Trigonometric equations - formulas, solutions, examples. Expressions using sine and cosine

>> Arctangent and arccotangent. Solving the equations tgx = a, ctgx = a

§ 19. Arctangent and arccotangent. Solving the equations tgx = a, ctgx = a

In Example 2 of §16 we were unable to solve three equations:

We have already solved two of them - the first in § 17 and the second in § 18, for this we had to introduce the concepts arc cosine and arcsine. Consider the third equation x = 2.
The graphs of the functions y=tg x and y=2 have infinitely many common points, the abscissas of all these points have the form - the abscissa of the point of intersection of the straight line y = 2 with the main branch of the tangentoid (Fig. 90). For the number x1, mathematicians came up with the designation acrtg 2 (read “arctangent of two”). Then all the roots of the equation x=2 can be described by the formula x=arctg 2 + pk.
What is agctg 2? This is the number tangent which is equal to 2 and which belongs to the interval
Let us now consider the equation tg x = -2.
Function graphs have infinitely many common points, the abscissas of all these points have the form abscissa of the point of intersection of the straight line y = -2 with the main branch of the tangentoid. For the number x 2, mathematicians came up with the notation arctg(-2). Then all roots of the equation x = -2 can be described by the formula

What is acrtg(-2)? This is a number whose tangent is -2 and which belongs to the interval. Please note (see Fig. 90): x 2 = -x 2. This means that arctg(-2) = - arctg 2.
Let us formulate the definition of arctangent in general form.

Definition 1. arсtg a (arc tangent a) is a number from the interval whose tangent is equal to a. So,

We are now in a position to draw a general conclusion about the solution equations x=a: the equation x = a has solutions

We noted above that arctg(-2) = -agctg 2. In general, for any value of a the formula is valid

Example 1. Calculate:

Example 2. Solve equations:

A) Let’s create a solution formula:

We cannot calculate the value of the arctangent in this case, so we will leave the solution of the equation in the form obtained.
Answer:
Example 3. Solve inequalities:
Inequalities of the form can be solved graphically, adhering to the following plans
1) construct a tangent y = tan x and a straight line y = a;
2) select for the main branch of the tangeisoid the interval of the x axis on which the given inequality is satisfied;
3) taking into account the periodicity of the function y = tan x, write the answer in general form.
Let us apply this plan to solve the given inequalities.

: a) Let’s construct graphs of the functions y = tgх and y = 1. On the main branch of the tangentsoid they intersect at the point

Let us select the interval of the x axis on which the main branch of the tangentoid is located below the straight line y = 1 - this is the interval
Taking into account the periodicity of the function y = tgх, we conclude that the given inequality is satisfied on any interval of the form:

The union of all such intervals represents the general solution to the given inequality.
The answer can be written in another way:

b) Let's build graphs of the functions y = tan x and y = -2. On the main branch of the tangentoid (Fig. 92) they intersect at the point x = arctg(-2).

Let us select the interval of the x axis on which the main branch of the tangentoid

Consider the equation with tan x=a, where a>0. The graphs of the functions y=ctg x and y =a have infinitely many common points, the abscissas of all these points have the form: x = x 1 + pk, where x 1 =arccstg a is the abscissa of the point of intersection of the straight line y=a with the main branch of the tangentoid (Fig. .93). This means that arcstg a is a number whose cotangent is equal to a and which belongs to the interval (0, n); on this interval the main branch of the graph of the function y = сtg x is constructed.

In Fig. 93 also presents a graphical illustration of the solution to the equation c1tg = -a. The graphs of the functions y = сtg x and y = -а have infinitely many common points, the abscissas of all these points have the form x = x 2 + pk, where x 2 = агсстg (- а) is the abscissa of the point of intersection of the straight line y = -а with the main line tangentoid branch. This means that arcstg(-a) is a number whose cotangent is equal to -a and which belongs to the interval (O, n); on this interval the main branch of the graph of the function Y = сtg x is constructed.

Definition 2. arccstg a (arc cotangent a) is a number from the interval (0, n) whose cotangent is equal to a.
So,

Now we are able to draw a general conclusion about the solution of the equation ctg x = a: the equation ctg x = a has solutions:

Please note (see Fig. 93): x 2 = n-x 1. This means that

Example 4. Calculate:

A) Let's say

The equation сtg x=а can almost always be converted to the form. An exception is the equation сtg x =0. But in this case, taking advantage of the fact that you can go to
equation cos x=0. Thus, an equation of the form x = a is not of independent interest.

A.G. Mordkovich Algebra 10th grade

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Topic: Trigonometric equations

Lesson: Arctangent and solving the equation tgx=a (continued)

1. Lesson topic, introduction

In this lesson we will look at solving the equation for any real

2. Solution of the equation tgx=√3

Problem 1. Solve the equation

Let's find the solution using function graphs (Fig. 1).

Let's consider the interval. On this interval the function is monotonic, which means it is achieved only for one value of the function.

Answer:

Let's solve the same equation using the number circle (Fig. 2).

Answer:

3. Solution of the equation tgx=a in general form

Let us solve the equation in general form (Fig. 3).

On the interval the equation has a unique solution

Smallest positive period

Let us illustrate on the number circle (Fig. 4).

4. Problem solving

Problem 2. Solve the equation

Let's change the variable

Problem 3. Solve the system:

Solution (Fig. 5):

At a point, the value is therefore the solution to the system is only the point

Answer:

Problem 4. Solve the equation

Let's solve using the variable change method:

Problem 5. Find the number of solutions to the equation on the interval

Let's solve the problem using a graph (Fig. 6).

The equation has three solutions on a given interval.

Let's illustrate it on a number circle (Fig. 7), although it is not as clear as on the graph.

Answer: Three solutions.

5. Conclusion, conclusion

We solved the equation for any real using the concept of arctangent. In the next lesson we will become familiar with the concept of arc tangent.

References

1. Algebra and beginning of analysis, grade 10 (in two parts). Textbook for general education institutions (profile level), ed. A. G. Mordkovich. -M.: Mnemosyne, 2009.

2. Algebra and beginning of analysis, grade 10 (in two parts). Problem book for educational institutions (profile level), ed. A. G. Mordkovich. -M.: Mnemosyne, 2007.

3. Vilenkin N. Ya., Ivashev-Musatov O. S., Shvartsburd S. I. Algebra and mathematical analysis for grade 10 (textbook for students of schools and classes with advanced study of mathematics). - M.: Prosveshchenie, 1996.

4. Galitsky M. L., Moshkovich M. M., Shvartsburd S. I. In-depth study of algebra and mathematical analysis. - M.: Prosveshchenie, 1997.

5. Collection of problems in mathematics for applicants to technical colleges (edited by M. I. Skanavi). - M.: Higher School, 1992.

6. Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebraic simulator.-K.: A.S.K., 1997.

7. Sahakyan S. M., Goldman A. M., Denisov D. V. Problems in algebra and principles of analysis (a manual for students in grades 10-11 of general education institutions). - M.: Prosveshchenie, 2003.

8. Karp A.P. Collection of problems on algebra and principles of analysis: textbook. allowance for 10-11 grades. with depth studied Mathematics.-M.: Education, 2006.

Homework

Algebra and beginning of analysis, grade 10 (in two parts). Problem book for educational institutions (profile level), ed. A. G. Mordkovich. -M.: Mnemosyne, 2007.

№№ 22.18, 22.21.

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