American Journal of Pedagogical and Educational Research ISSN (E): 2832-9791| Volume 10, | Mar., 2023 P a g e | 147 www.americanjournal.org TEACHING STUDENTS TO SOLVE TRIGONOMETRIC EQUATIONS USING ELEMENTS OF MATHEMATICAL ANALYSIS IN MATHEMATICS CIRCLES Turginov A. M. Q.D.P.I Senior Teacher Askaraliyeva M. A. Q.D.P.I Teacher A B S T R A C T K E Y W O R D S Mathematics circles, mathematics is the main type of work outside the classroom. Circles arouse students' interest in science and improve their mathematical thinking, skills, and the quality of their mathematical training. This increases students' thinking range. Introduction There are the following methods of using elements of mathematical analysis in solving trigonometric equations. 1. Using the field of definition of the function 2. Use of the property of boundedness of the function 3. Using properties of sine and cosine functions 4. Avoiding numerical inequalities. 1. Using the field of definition of the function 1. Example tgxxx +−= 4 sinsin Solve equation (1). Solving. The field of definition of Eq consists of From this zkkx = , . x Putting this value of in equation (1), we see that its right and left sides are equal to 0. So everyone zkkx = ; will be the root of the equation. J: zkkx = ; . 2. Use of the property of boundedness of the function Example 2. 1log1)sin(cos 22 5 2 +++= xxxx Solve equation (2). Solution: the equation is defined for all real X's. For an arbitrary X        + −  znnx x x ; 2 0sin 0sin   American Journal of Pedagogical and Educational Research Volume 10, Mar., 2023 P a g e | 148 www.americanjournal.org 11log1,1)sin(cos 22 5 2 +++ xxxx As a result, equation (2) is equivalent to the following system of equations. the general solution x=0 follows from this system. Answer: x=0 Example 3. Solve the equation. ( ) ( )x x xx + ++= + 5log 1 5log 6 cos2 5 5 2 2 Eat: ( )ye according to ( ) ( ) ( ) ( ) 0 2 5log 1 5log 2 6 cos2 2 5log 1 5log 2 6 cos2 5 5 2 2 5 5 2 2 =        = + ++ = +          + ++  + x x x xx x x xx it follows that Answer: x=0 3. Using properties of sine and cosine functions Example 4 27cos3cos 113 −=+ xx solve the equation. Solution: If 0x is a solution of the equation, then 13cos 0 −=x (otherwise 17cos 0 −x it cannot be). So, 17cos 0 −=x As a result, the arbitrary solution of the equation will be the solution of the following system.    −= −= 17cos 13cos x x arbitrary solution of the system is the solution of the equation. Therefore, the equation is as strong as the above system. The first equation of the system zk k xk += , 3 2 3  has a solution. From these solutions, we find the ones that satisfy the system equation 2. These satisfy the following equality zm are numbers. m k   2 3 14 3 7 +=+ we write in the following form: 7 23 − = m k Equality since K and M are integers zttm += ,37 is appropriate, but in this zttk += ,13 So, this is the solution of the system kx that .,13 zttk += zttx ++= , 3 2 2 3    So the general answer is: zttx += ,2 4. Avoiding numerical inequalities.     =++ = 01log 1)sin(cos 22 5 2 xx xx American Journal of Pedagogical and Educational Research Volume 10, Mar., 2023 P a g e | 149 www.americanjournal.org Example 5. ( ) 2 2 228 28 4 cos42cossin 2cos 1 sin 1 xxx xx −=+      +  solve the equation.Solution: For arbitrary positive numbers and 4)( 11 +      + ba ba we know that inequality is reasonable. In the field of definition of Eq 02cos,0sin 28  xx using the inequality, we see that the left side of the equation is not less than 4. At the same time, in the field of definition of Eq 4 4 cos0 2 2 2 − x  As a result, the equation is equivalent to the following system of equations. ( )        =− =+      + 1 4 cos 42cossin 2cos 1 sin 1 2 2 2 28 28 x xx xx  So, 2 1  =x and 2 2  −=x will be the solutions of the given equation. List of Used Literature 1. И.Х. Сивашинский «Элементарные функции и графики» Москва. “Наука” 1968г.[95,134] 2. T. Karimov, K. To‘raqulov. S. Abdullayev. “Maktabda tenglama va tengsizliklar” Toshkent. “O‘qituvchi”, 1992 yil.[54,86] 3. Turg’inov A.M, Asqaraliyeva M.A Matematika to’garaklarida o’quvchilarga ba’zi tenglamalarni tenglamalar sistemasiga keltirib еchishni o’rgatish. 4. Turginov Azizjon Mamasaliyevich, Asqaraliyeva Muktasarkxon Azizjon kizi A non –local problem for a third-order equation with multiple chrarasteristics Vol.42,Issue-05,2022 pp.230-241 5. Тургинов А.М., Асқаралиева М.А. об одной нелокальной задаче для уравнения третьего порядка с нелокальным условиям. IJSSIR, Vol. 11, No. 06. June 2022, с 66-73.