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AP EAMCET201823 Apr 2018Morning ShiftMathematicsTrigonometric EquationsActual

The number of solutions of the equation 4 2 3 = , when 0 < < , is

Options

  1. A2
  2. B4
  3. C6
  4. 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 .

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