Tiêu đề XI - maths - chapter 3 - TRIGONOMETRIC EQUATIONS (WE_Level-5_6) (131-161).pdf ✅

Narayana Junior Colleges TRIGONOMETRIC EQUATIONS JEE ADVANCED - VOL - II Narayana Junior Colleges 131 TRIGONOMETRIC EQUATIONS WORKEDOUT EXAMPLES: W.E-1:Solutions of the equation 4 4 7 sin cos sin cos 2 x x x x   . Ans:  1 2 12 n n x where n Z       Sol: 4 4 4 2 2 4 2 2 sin cos sin 2 sin cos cos 2 sin cos x x x x x x x x        2 2 2 2 2    sin cos 2sin cos x x x x where we get   4 4 2 1 sin cos 1 sin 2 ...... 2 x x x i    Using (i), the given equation becomes 1 7 2 1 sin 2 sin 2 2 4   x x 2 2sin 2 7sin 2 4 0 (2sin 2 1)(sin 2 4) 0 x x x x        either 2sin 2 1 0 2 1   6 n x x n where n Z          or sin 2 4 0 x   which has no solution because - 4 range of sin 2x where n Z  So, the solution of equation is  1 2 12 n n x where n Z       . W.E-2:Solutions of the equation 3 3 3 3 sin cos 3 3 sin cos 1 x x x x    . Ans: 2 1 2 ; ,  3  x k x m m I k I          Sol: Let a x b x c     3 sin ; cos ; 1 then given equation reduces to 3 3 3 a b c abc     3 0    2 2 2          a b c a b c ab bc ca 0     2 2 3sin cos 1 3sin cos 1 3sin cos cos 3sin 0 x x x x x x x x          . Thus the given equation has solution if 3 sin cos 1 0 x x    3 1 1 3 sin cos 1 sin cos 2 2 2 sin sin 6 6 x x x x x                   1 6 6 n x n               . If n is odd; 2 1 6 6 3 3 x n n x k                  If n is even 2 ; 6 6 x n n x m m I              . W.E-3: Solution set of the equation 2 2     6 6 10 10 log sin 3 sin log sin 2 x x x x x x x       . Ans: 5 3   Sol: 2 2     6 6 10 10 log sin 3 sin log sin 2 x x x x x x x       sin3 sin 0 x x   and sin 2 0 x  2sin 2 cos 0 x x and  st x I  quadrant. 2 6 10 sin 3 sin log 0 sin 2 x x x x x           or 2sin 2 cos 1 sin 2 x x x  or 2cos 1 x 
Narayana Junior Colleges JEE ADVANCED - VOL - II TRIGONOMETRIC EQUATIONS 132 Narayana Junior Colleges or cos 1/ 2 2 / 3 x x n         2 2 6 6 6,0 0 1 10 10 x x x x x and                 . for n=-1 x= 5 3 -2 3      π π π W.E-4: Solution of the equation sin cos 2 sin 2 2 x x x               Ans:     2 2 1 1 1 6 2 n n x n or x n                        Sol: sin cos 2 sin 2 2 x x x               2 1 2sin cos 2sin 2 2 x x   x 2 1 sin 2sin   x x 2 2sin sin 1 0 x x    2 2sin 2sin sin 1 0 x x x     2sin sin 1 1 sin 1 0 x x x         1 sin sin 1 2 x or x     1 ; 1    6 2 n n x n x n                      2 2 1 1 1 6 2 n n x n or x n                        W.E-5: The least positive angle measured in degrees satisfying the equation   3 3 3 3 sin sin 2 sin 3 sin sin 2 sin 3 x x x x x x      Ans: 0 72 Sol:   3 3 3 3 sin sin 2 sin 3 sin sin 2 sin 3 x x x x x x              3 3 3 3  a b c a b c a b b c c a          0      sin sin2 sin2 sin3 sin3 sin 0 x x x x x x    3 5 2sin cos 0 ( ) 2sin cos 0 2 2 2 2    x x x x or (or) 2sin 2 cos 0 x x  3 2 sin 0 ; 2 3 x n x n Z       or 5 2 sin 0 ; 2 5 x m x m Z      or sin 2 0 ; 2 n x x k Z      or cos 0 2 1 ;   2 x      x p p Z  or cos 0 2 1 ;   2 x x q q Z       .  Least positive angle is 2 0 72 5   . W.E-6:The set of value of ' ' a for which the equation 4 4 sin cos sin 2 0 x x x a     possesses solutions. Ans: 3 1, 2 2 a        Sol: 4 4 sin cos sin 2 0 x x x a     .     2 2 2 2 2 2 2 2 sin cos 2sin cos 2sin cos 2sin cos 0 x x x x x x x x a       2 sin 2 1 sin 2 0 2 x     x a 2 2 sin 2 2sin 2 2 0     x x a or 2 2 sin 2 2sin 2 2 a x x    or   2 2 3 sin 2 1 a x        2       2 3 1 1,1 a y where y    2 3 0,4 a   3 1, 2 2 a          sin 2 1 2 3 x a   
Narayana Junior Colleges TRIGONOMETRIC EQUATIONS JEE ADVANCED - VOL - II Narayana Junior Colleges 133 sin 2 sin x   where   1  sin 1 2 3 a     or     1 1 1 sin 1 2 3 2 n x n a            where 3 1, 2 2 a        . W.E-7:The general solution of the equation, 2 1 2 1 2 1 2 sin sin 3cos 0 3 3 x x x x x x       . Ans:   1 3 1 1 sin 3 2 4 n x where n                Sol: 2 1 2 1 2 1 2 sin sin 3cos 0 3 3 x x x x x x         2 4 2 1 2 1 2 1 2 sin co s 3 co s 2 3 3 3      x x x x x x or   2 2 2 1 2 1 2 1 2sin cos 3cos 3 3 3 x x x x x x     or   2 2 2 1 2 1 2 1 4sin cos 3cos 0 3 3 3 x x x x x x            2 2 1 cos 0 3 x x          (or) 2 1 sin 3 x x    3 3 1 sin , 1 sin 4 4                n where n 2 2 3 1 cos cos 3 2 x x          2 1 3 2 x n x         1 3 2 1 2 x n      or 2 1 sin sin 3 x x            1 3 1 1 sin 3 2 4 n x where n                 W.E-8:Solve the equation             2 cos 3 2cos 3 2cos 4 3 cos 7 3 x x x x                      2 1 sin 3 2 sin 3 2 sin 4 3 2 sin 3 sin 7 3 x x x x x            Ans: log 1 3   4 16 16 n n x             , n Z  Sol: Let 3x    then 2 2 cos 2cos 2cos4 cos7 sin 2sin 2sin 4 2sin3 sin7                     cos cos 7 sin 7 sin 2 cos 4 sin 4 2 2 sin 3                  2sin 4 sin 3 2cos 4 sin 3 2 cos 4 sin 4 2 2sin 3                    s in 4 s in 3 1 c o s 4 s in 3 1 1 s in 3 0                  sin 3 1 sin 4 cos 4 1 0      sin 3 1 3 2 2 n           3 3 2 / 2 x       n 1 3 1 4 1 3 2 1 log 2 2 x n n x         3 4 1 log 1 2 n x     And sin 4 cos 4 1     sin 4 sin 4 1   4 4 4 4 n n                      4 3 1   4 4 n x n          3 1   4 16 16 n x n       
Narayana Junior Colleges JEE ADVANCED - VOL - II TRIGONOMETRIC EQUATIONS 134 Narayana Junior Colleges log 1 3   4 16 16 n n x             , n Z  W.E-9: The value of ' ' a so that the equation     2 2 2 2 cos 2cos 5cos 2 sec sec 10 x x x a x            has a real solution. Ans: 2 6 r a where r I      Sol: Since 2 cos 1   and 2 sec 1     2 2 cos 2cos 5cos 2 1 10 x x              2 2cos 5cos 2 2 10 x x n n I         2      2cos 5cos 2 20 , x x n n I . Now 2 2 5 9 2cos 5cos 2 2 cos 4 8 x x x            2       1 2cos 5cos 2 9 x x  In equation 1 only one value of n is possible i.e., n  0 2     2cos 5cos 2 0 x x     2cos 1 cos 2 0 x x   2 3 x n      2 2 sec 2 sec 2 1 3 3 n a n                      2 sec 2 2 1 3 n a             2 3 a r      2 6 r a      where r I  . W.E-10: The number of roots of the equation, sin 2sin 2 3 sin3 x x x    is [IIT 2014] (A) 0 (B) 1 (C) 2 (D) infinite Ans: A Sol: sin sin 3 2 sin 2 3 x x x         2sin cos 2 2sin 2 3 x x x     2sin cos 2 2sin 2 3 0 x x x     2 2 2 2 2sin cos2 2sin2 sin cos sin 2 cos 2 1 0 x x x x x x x             2 2 2       1 sin 2 sin cos 2 cos 0 x x x x Hence there is no root because all terms do not vanish simultaneously. W.E-11: Total number of solutions of sin cos 0 x x    for x0, 2  is (where . denotes the greatest integer function) (A) 1 (B) 2 (C) 4 (D) 0 Ans: D Sol: Given equation is sin cos x x    Real solution exist iff - cosx is integer i.e., cos x is an integer. So possible solutions for x0, 2  are x  0, / 2, ,3 / 2,2     . when x  0, 2 RHS = -1 and LHS = 0 no solution x or    / 2 3 / 2 RHS = 0 and LHS = 1 no solution x   RHS = 1 and LHS = 0 no solution. So, we can conclude that equations has no real solution exist. W.E-12: Number of solutions to the equation   sin 2sin sin 2sin 2 3 1 0, 2 x x x x e e e e in       is (A) zero (B) two (C) four (D) one Ans: (B) Sol: Let sin ' ' x e t  then 2 2 2 3 1 t t t t      2 2       t t t t 2 3 1 here 2 2 t t t t t R        2 3 0, 1 0 ,