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Question

Number of values of θ which satisfy sin4θ2sin2θ1=θ is?

A
0
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B
Infinite
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C
Equal to the number of values of θ satisfying 8sec2θ6secθ+1=0
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D
None of these
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Solution

The correct option is A 0

sin4θ2sin2θ1=0

Let sin2θ=y

y22y1=0

y=2±4+42=1±2

sin2θ=1+2 or sin2θ=12

sin2θ=1+2 will give value of

sin θ greater than

1 which is impossible

sin2θ=12 is also impossible because sin2θ cannot be negative.

there is no possible θ for above given trigmometric equation

. Answer: option: (A)

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