Tiêu đề XI - maths - chapter 3 - TRIGONOMETRIC EQUATIONS (70-89).pdf ✅

TRIGONOMETRIC EQUATIONS JEE-MAINS ADVANCED  Equations involving one or more than one trigonometric ratios of unknown angles are called trigonometric equations. Ex: 2 3 4 0 C os C os     is trigonometric equation in unknown angle ' '.  Solution of a trigonometric equation : A value of the unknown angle satisfying the given equation is called the solution of the given equation. Every other coterminal angle of a solution is also solution of the given trigonometric equation.  The trigonometric equation may have infinite number of solutions and can be classified as (a) Principal solution (b) General solution (a)Principal solution: If  has infinitely many solutions among them, the least value with sign is called principal value of  . It is denoted by ' ' . (i) There exists a unique value of ' '  in , 2 2         satisfying sin , , 1     k k R k . This value of ' '  is called principal value of ' '  or prinicpal solution of sin  k (ii) There exists a unique value of ' '  in 0,  satisfying cos , , 1     k k R k . This value of ' '  is called principal value of ' '  or principal solution of cos  k (iii) There exists a unique value of ' '  in , 2 2          satisfying tan ,    k k R . This value of ' '  is called principal value of ' '  or principal solution of tan  k . (b) General solution: Since trigonometric functions are periodic, a solution generalised by means of periodicity of trigonometric functions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution. S n o Equation Interval in which principal solution  General Solution 1   sin , 1 1 k k      , 2 2          1 , n n n Z        2   cos , 1 1 k k      0,  2n , n Z      3   tan k, k R    , 2 2         n , n Z      4 cos , ( , 1] [1, ) ec k k        , 0 2 2           1 , n n n Z        5 sec , ( , 1] [1, ) k k        0,  2          2n , n Z      6 cot k, k R    0,  n , n Z       The general solution of 2 2 sin sin    (or) 2 2 cos cos    (or) 2 2 tan tan    is       n n Z ,  Common Solution of Two Equations: The general solution for the equations of the type sin sin    , cos cos    is      2n where,  is common angle lies between 0 and 2 W.E-1:The solution of 1 sin 2    , 1 tan 3   is Sol: sin 0   and 3 tan 0     Q , principal common solution is      / 6 7 / 6  general solution 7 2n , n Z 6        Some important values to remember: Angle Sin Cos 150 2 2 3 1 2 2 3 1 180 4 5 1 10 2 5 4  0 2 1 22 2 2 2  1 2 2 2  1 360 10 2 5 4  4 5  1 TRIGONOMETRIC EQUATIONS SYNOPSIS
JEE MAINS - C.W - VOL - II MATHS BY Er. RG SIR TRIGONOMETRIC EQUATIONS 540 4 5 1 10 2 5 4  0 2 1 67 2 2 2 1 2 2 2 1 720 10 2 5 4  4 5  1 750 2 2 3  1 2 2 3  1  i) tan 150 = 2  3 , tan 0 2 1 22 = 2 1, ii) cot 150 = 2  3 , cot 0 2 1 22 = 2  1  i) The equation a b c cos sin     has a solution if 2 2 c a b   . ii) The equation a b c cos sin     has no solution if 2 2 c a b    Important Points to Remember: (i) For equations of the type sin = k or cos = k ,one must check that k 1.  (ii) Avoid squaring the equation, if possible, be cause it may lead to extraneous solutions. (iii) Do not cancel the common variable factor from the two sides of the equations, because we may loose some solutions. (iv) The answer should not contain such values of  which make any of the terms undefined or infinite. (v) Check that denominator is not zero at any stage while solving equations. (vi) (a) If tan  or sec is involved then  should not be an odd multiple of  / 2 . (b) If cot  or cosec  is involved then  should not be a multiple of  or 0 (vii) If two different trigonometric ratios, such as, tan and sec are involved then after solving we cannot apply the usual formulae for general solution, because periodicity of the functions are not same. (viii) If L.H.S.of the given trigonometric equation is always less than or equal to k and R.H.S. is always greater than or equal to k.If both the sides are equal to k for same value of  ,then solution exist and if they are equal for different value of ,then solution does not exist.  Solution of Inequations: If a < b, then             (i) x a x b 0 a x b (ii) x a x b 0 a x b (iii) x a x b 0 x a or x b (iv) x a x b 0 x a or x b.                         PRINCIPAL SOLUTION 1. If 4 3 2 Cos   then       1) 6 5 , 6   2) 4 3 , 4   3) 3 2 , 3   4) 2   2. If Tan  Sec  3 , then the principal value of 6         is 1) 3  2) 4  3) 3 2 4) 2  3. If Tan  3  3 1Tan 2    and  lies in        2 , 2   then  = 1) , 3 4         2) , 3 4         3) , 6 3         4) , 12 12         4. If              4 4   Tan Cos Cot Sin and  is in the first quadrant then  =----- 1) 3  2) 2  3) 4  4) 6  5. If 3 10 3 0 4 2 Tan   Tan    then principal values of ‘ ’ are 1) 0 0  45 ,36 2) 0 0  30 ,60 3) 0 0  75 ,36 4) 0 0  60 ,15 6. If     0 0 Sin x  28  Cos 3x  78 then x = 1) 370 , 80 2) 390 3) 350 , 80 4) 470 7. If sin 2  cos3  0 and  is acute then  = 1) 360 2)540 3)180 4) 300 8. The smallest value of  satisfying the equation 3Cot Tan   4 is 1) 3 2 2) 3  3) 6  4) 12  LEVEL - I
TRIGONOMETRIC EQUATIONS JEE-MAINS ADVANCED GENERAL SOLUTION 9. The most general value of  satisfying the equations 2 1 Sin  , 2 1 Cos   is 1) 4 2  n  , n Z 2) 4 3 2  n  , n Z 3) 6  n  , n Z 4) 3  n  , n Z 10. The general solution of 2 1 1 . 2 Tan x Tanx Tanx Tan x    is 1) 4  n  , n Z 2) 4  n  , n Z 3)  4) 6  n  , n Z 11. If 11Sin2 x + 7Cos2 x = 8 then x =----- 1) 6  n  , n Z 2) 4  n  , n Z 3) 3  n  , n Z 4) 2  n  , n Z 12. The general solution of Sin2x=4Cosx is 1) 2 (2 1)  n  , n Z 2) n , n Z 3) Empty Set 4) 2n , n Z 13. 3 4 4 2 4 2                 A Tan A Tan   ; then the general solution of A = 1) 6 2  n  , n Z 2) 4  n  , n Z 3) 4 2  n  , n Z 4) n , n Z 14. If Tanx+2Tan2x+4Tan4x+8Cot8x= 3 then the general solution of x = 1) 3  n  , n Z 2) 6  n  , n Z 3) 4  n  , n Z 4) n , n Z 15. The general value of  satisfies the equation     3 1 120 120 0 0 TanTan  Tan   is 1)  18 6 1  n  , n Z 2)  3 3 1  n  , n Z 3)   6 6 1  n  , n Z 4)  6 3 1  n  , n Z 16. If Tanm  Cotn then the G.S of  = 1)   m n k   2 1 , k Z 2)   m n k   2 2 1 , k Z 3)   m n k  2 1  , k Z 4)   m n k  2 1  , k Z 17. If Cosx+Cosy=1 and CosxCosy=1/4 then the G.S are 1) x  n  n  Z y  m  ,m  Z 4 , , 2 4 2     2) x  n  n  Z y  m  ,m  Z 3 , , 2 3 2     3) x  n  n  Z y  m  ,m  Z 6 , , 2 6 2     4) x  n  n  Z y  m  ,m  Z 5 , , 2 5 2     18. If     0 0 1 . 60 . 60 8 Sin x Sin x Sin x    , then x = 1)   6 1   n n   2)   18 1 3 n n    3)   3 1   n n   4)   9 1 3 n n    SOLUTIONS IN THE GIVEN INTERVAL 19. The value of  satisfying Sin7  Sin4  Sin in 2 0     are 1) 4 , 9   2) 9 , 3   3) 4 , 6   4) 0,1 20. If 1 Sin  3SinCos 2   then the solution set in       2 0,  is 1) 1 1 , 4 3 Cos         2) 1 1 , 4 2 Tan         3) 1 1 , 3 3 T an         4) 1 1 , 6 3 Sin         21. The equation 3Sinx  cosx  4 has 1) Only one solution 2) Two solutions 3) Infinitely many solutions 4) No solution 22. If y  Cos  Sin has a real solution then 1)  2  y  2 2) y  2 3) y   2 4) y  1
JEE MAINS - C.W - VOL - II MATHS BY Er. RG SIR TRIGONOMETRIC EQUATIONS LEVEL-I-KEY 01) 1 02) 1 03) 2 04) 3 05) 2 06) 3 07) 3 08) 3 09) 2 10) 3 11) 1 12) 1 13) 1 14) 2 15) 1 16) 2 17) 2 18) 2 19) 1 20) 2 21) 4 22) 1 LEVEL-I-HINTS 1. 2 4 cos 3                  2 2 3 cos n 2 6 2. tan sec 3 sin 1 3 cos          3 1 1 3 cos sin 1 cos sin 2 2 2           1 1 cos cos sin sin cos 6 6 2 6 2                   P.S. of 6 3      3.                  2 tan 3 3 1 tan , 2 2 on verification     , 3 4 satisfies 4.                  c o t t a n sin c o s 4 4 on verification 4    satisfies 5. 4 2 3 tan 10 tan 3 0      4 2 2      3 tan 9 tan tan 3 0             2 2 3 tan 1 tan 3 0           2 2 1 1 tan 3 3   2 2 ; tan 3 3    0 0      30 , 60 6.   0 0 0 3x 78 90 x 28     or 90+x+280        0 0 0 0 0 0 140 4x 90 28 78 140 x 35 4 x = 80 7. sin 2 cos 3    0 0 3 2 5 90 18 2            8. 4 2cosec2 = 3  3 sin 2 2 2 3       ; 6    9. 1 1 sin ; cos 2 2        IIquadrant         3 P.S. of 4 4 G.S. of 3 2n , V n Z 4       10. tan 3x 2x 1 tan x 1 x   4        But for   x , 2x tan 2x    4 2 is undefined  no solutions 11. 2 2 2 11 sin x 7 cos x 8 4 sin x 7 8      2 2 2 1 1 4 sin x 1 sin x 4 2            x n 6      12. sin 2x 4 cos x 2 sin x cos x 4 cos x       cos x 0 or sin x 2 (which is not possible), x 2n 1 , n Z   2      13. A A 4 tan tan 4 2 4 2 3                   A A 4 tan cot 4 2 4 2 3                                            A 4 2cosec 2 tan cot 2cosec2 4 2 3 4 2 3 2cosec A secA cosA 2 2 3 3               A 2n , Vn Z 6       14. tan 2 tan 2 4 tan 4 8 cot 8 3 x x x x     cot x 8 cot 8x 8 cot 8x 3 cot x 3      x n , n Z 6      