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Q.

The values of  θ satisfying  sin5θ=sin3θsinθand  0<θ<π2 are

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a

π6,π4

b

π4,π3

c

π6,π3

d

π4,π2

answer is A.

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Detailed Solution

sin5θ=sin3θsinθ 0θ<π2

sin5θ+sinθ=sin3θ

2sin3θcos2θ=sin3θ

sin3θ(2cos2θ1)=0

sin3θ=0,2cos2θ=1

3θ=nπ (or) cos2θ=cos60°

θ=nπ3 (or) 2θ=2nπ±π3

θ=nπ±π6

0<θ<π2,  θ=π6,π3 

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