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Question

The general solution of the equation sin3θ cosθcos3θ sinθ=14 is

A

nπ4+(1)n+1(π4), nZ
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B

nπ4+(1)n+1(π8), nZ
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C

nπ8+(1)n+1(π4), nZ
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D

nπ8+(1)n+1(π8), nZ
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Solution

The correct option is B
nπ4+(1)n+1(π8), nZ
Here, we can write the equation as:
sin3θ cos θcos3θ sin θ=144sin θcos θ(sin2θcos2θ)=12sin 2θ(sin2θcos2θ)=12sin 2θ(cos2θsin2θ)=12sin 2θcos 2θ=1sin 4θ=1sin 4θ=sin(π2)4θ=nπ+(1)n(π2), nZ4θ=nπ+(1)n+1(π2), nZθ=nπ4+(1)n+1(π8), nZ
Hence, Option b. is correct.

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