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A value of θ0,π3 for which 1+cos2θsin2θ4cos6θcos2θ1+sin2θ4cos6θcos2θsin2θ1+4cos6θ=0 is

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a
7π36
b
7π24
c
π18
d
π9

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detailed solution

Correct option is D

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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