11 AMERICAN Journal of Public Diplomacy and International Studies www. grnjournal.us AMERICAN Journal of Public Diplomacy and International Studies Volume 01, Issue 02, 2023 ISSN (E): ХХХ-ХХХ Non-Standard Methods for Solving Trigonometric Inequalities Sh. L. Ermatov Andijan State University Abstract: This article describes non-standard ususes for solving trigonometric equations. It presents the method of sectors and the method of concentric circles from non-standard methods of solving trigonometric inequalities based on the principle of the genetic approach. Keywords: trigonometric issue, method of concentric circles, method of sectors, genetic approach. Introduction: The purpose of teaching trigonometry in specialized schools (educational, educational, practical) is determined by the implementation of trigonometry in solving practical issues related to various fields and disciplines, as well as the application of trigonometric functions. Trigonometry is distinguished by great importance in theory and practice, in particular, by the fact that it is an excellent computational apparatus in solving various problems of geometry, by the means of trigonometric functions it is convenient to express mathematical concepts such as periodicity, monotony, pointing accuracy, delimitation, pairwise rigidity, bias and convex, continuity in a clear, visual and understandable way [2,3,4]. In particular, the use of non-standard methods in solving inequalities of the trigonometric equation extends the field of application of trigonometry. Below we give several non-standard methods for solving trigonometric inequalities. Research Methodology: a) Sectors method in solving trigonometric inequalities. Let's consider the method of sectors when solving trigonometric inequalities. in the form (where and rationally introduced trigonometric functions) the solution of rational trigonometric inequalities is almost identical to the solution of rational inequalities. It is convenient to solve rational inequalities by the method of intervals on the number axis. As its analogue and for the trigonometric circle, and for the solution of the rational trigonometric inequality in the semicircle, the method of sectors is considered [5,6,7]. 1. solving form inequalities. ( ) 0, ( 0, 0, 0) ( ) P x Q x     ( )P x ( )Q x sin x cosx ( 2 )T  tgx ctgx ( )T  (sin ,cos ) 0 ( 0, 0, 0) (sin ,cos ) P x x Q x x     12 AMERICAN Journal of Public Diplomacy and International Studies www. grnjournal.us The numerator and denominator of an expression given in the interval method are whose factors correspond to it when passing through the point, the exchange of signals is determined, and the corresponding intervals are obtained depending on the signal of the inequality. In the sectoral method, however, each factor of the numerator and denominator of a given expression is in form. Here - or is one of the functions, in the trigonometric circle and divides two sectors corresponding to the corners. These corners are is equivalent to from in the transition to the signal of the expression is determined. The following should be borne in mind: a) and form factors while does not change the signal at all values of. Therefore, such multipliers are discarded. If if, by changing the inequality signal to the opposite, the factors are discarded. b) and form factors are also discarded. Moreover, if these are multiples of the denominator, then and will be thrown under the conditioni. or when form factors are discarded, the inequality signal changes to the opposite. 2. solving form inequalities. or was one of the functions of the function each factor of the form is a trigonometric semicircle (or ) at one fits the angle, or . This when passing a point the signal of the multiplier alternates. In addition, function is not defined at, and will have a different signal on the left and right of these points. Similar function va at undefined and to the right of these points there are different symbols. 1- example. Solve the inequality: . Solution: . Trigonometric half in the circle the equation is given by angle, the equation is given by the angle is suitable. They divide the semicircle into three sectors, and their in each the function saves the signal and exchanges hints in each sector (1-picture). 1-picture 0( )x x 0x ( ( ) )f x a ( )f x sin x cos ,x 1 1a   1x 2x 1 2( ( ) ( ) )f x f x a  1x 2x ( ( ) )f x a (sin )x a (cos )x a | | 1a  x 1a  (sin 1)x  (cos 1)x  sin 1x   cos 1x   (sin 1)x  (cos 1)x  ( , ) 0 ( 0, 0, 0) ( , ) P tgx ctgx Q tgx ctgx     tgx ctgx ( )f x ( ( ) )f x a 2 2 x      0 x   0x 0( )f x a 0x 0( ( )f x a tgx 2 x    ctgx 0x  x  2 0ctg x ctgx  ( 1) 0ctgx ctgx   0 x   0ctgx  1 2 x   1ctgx   1 3 4 x   ( 1)y ctgx ctgx  13 AMERICAN Journal of Public Diplomacy and International Studies www. grnjournal.us in the sector . And in the rest, the hint alternates. period of function . since it is positive-signal sectors we choose. Answer: . b) method of concentric circles solving a system of trigonometric inequalities. This method is compared to the parallel number axis method when solving a system of rational inequalities. Let's see the following example. If the system of inequalities is solved on a single number of axes, the image presents difficulties in isolating the solution, it is much easier to isolate the solution if solved on the parallel number axis (2-picture). 2-picture. Solving inequalities on the parallel number axis. Answer: . 1- example. Solve the system of inequalities: 0 2 x    ( 1) 0ctgx ctgx   ( 1)y ctgx ctgx  T  2 0ctg x ctgx  2 , 3 4 n x n n Z n x n                   3 2 ( 2) ( 1)( 4) 0, ( 4) ( 2) 0, ( 3)( 3) ( 1)( 2) 0. ( 5)( 4) x x x x x x x x x x x x x                   { 2} [ 1;0) (4; )   14 AMERICAN Journal of Public Diplomacy and International Studies www. grnjournal.us Solution. We solve each inequality separately in the unit circle. 1) Argument is solving the inequality (3-picture). 3-picture. 2) and we solve for inequalities (4-picture). 4-picture. Now for arguments, we draw concentric circles. We draw a circle corresponding to the solution of the first inequality and we are the shitrich, then we draw a circle with a larger radius and we are the shitrich according to the solution of the second, then we draw a circle and a base circle for the third inequality. From the center of the circle, we pass rays through the ends of the arcs so that they cross all circles. The result is a solution in the base circle (5-picture). 1 cos , 2 3 sin 2 , 2 1 x x tgx            2x 3 sin 2 2 x  1 cos 2 x   1tgx   x 15 AMERICAN Journal of Public Diplomacy and International Studies www. grnjournal.us 5-picture. Answer: . Example 2. Solve the inequality: . Solution. using the formula, we substitute the form: , , . We solve for the last inequality. let's enter a mark, as a result: . We describe the solution in the unit circle (6-picture): . let's go back to the variable: . 3 7 2 ; 2 , 4 6 k k k Z             22sin 3cos2 0 4 x x         21 cos2 2sin   1 cos 2 3cos2 0 2 x x          cos 2 3cos2 1, 2 x x           sin2 3cos2 1x x   1 3 1 sin 2 cos2 , 2 2 2 x x   1 sin sin2 cos cos2 , 6 6 2 x x      1 cos 2 6 2 x         2 6 t x    1 cos 2 t   2 2 2 2 3 3 k t k         x 2 2 2 2 2 3 6 3 k x k            5 , 4 12 k x k k Z          16 AMERICAN Journal of Public Diplomacy and International Studies www. grnjournal.us 6-picture. Conclusion/Recommendations: In conclusion, high requirements are imposed on the knowledge, skills and qualifications of students of a specialized school, of which, first of all, it is required to have a high level of admission indicators to higher educational institutions, while maintaining a deep and fluent mastery of educational material in order to achieve positive results in Olympiads conducted in different disciplines. In this case, the application of non-standard methods opens the way for the development of methodological issues of teaching mathematics and the solution of problems in its teaching. References: 1. Sh.L. Ermatov. Trigonometrik tenglamalar yechimlari to’plamini umumlashtirishning bir usuli. Наманган давлат университети илмий ахборотномаси. -Наманган, 2021. -№ 10. - Б.445-450. 2. Khankulov, U. K. (2017). Description of Methodical System of Teaching Elements of Stochastics Line Mathematics Using Computer Technologies. Eastern European Scientific Journal, (6). 3. 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