AP EAMCET201823 Apr 2018Morning ShiftMathematicsTrigonometric EquationsActual
The number of solutions of the equation 4 2 3 = , when 0 < < , is
Options
- A2
- B4
- C6
- D8
Correct answer
C. 6
Step-by-step solution
We have, aligned & 4 2 3 = , (where 0 < < ) & 2(2 2 3 )= 1 & 2[ 5 + (- )]= 1 & [ 2 A B= (A+B)+ (A-B)] & 2[ 5 + ]= 1 [ (- )= ] & 2 5 +2 ^2 =1 & ( 6 + 4 )+ (2 ^2 -1 )=0 & 6 + 4 + 2 =0 & 2 4 2 + 4 =0 & 4 (2 2 +1)=0 & 4 =0 and 2 2 +1=0 & 4 =(2 n+1) 2 , 2 =- 1 2 & =(2 n+1) 8 , 2 =2 / 3 or 4 3 & = 8 , 3 8 , 5 8 , 7 8 , 3 , 2 3 aligned So, number of solution of given equation is 6 .