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Q.

The number of solutions of the equation sin4θ−2sin2θ−1=0 which lie between 0 and 2π is

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a

2

b

8

c

0

d

4

answer is A.

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Detailed Solution

sin4θ−2sin2θ+1=2

(sin2θ−1)2=2=cos4θ (not possible)

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