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A value of θ∈0,π3 for which 1+cos2θsin2θ4cos6θcos2θ1+sin2θ4cos6θcos2θsin2θ1+4cos6θ=0 is

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a

7π36

b

7π24

c

π18

d

π9

answer is D.

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Detailed Solution

We have R2→R2-R1, R3→R3-R2⇒1+cos2θsin2θ4cos6θ-110-101=0⇒1+cos2θ·1-sin2θ(-1)+4cos6θ=0⇒2+4cos6θ=0⇒cos6θ=-12⇒6θ=2π3,4π3⇒θ=π9,2π9

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