For n∈Z,2sin2θ=cos2θ implies θ2=

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For $n \in Z, 2 \sin^{2} θ = \cos 2 θ$ implies $\frac{θ}{2} =$

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2 n π \pm \frac{π}{3} n \in Z

\frac{n π}{2} \pm \frac{π}{12}, n \in Z

n π \pm \frac{π}{12}, n \in Z

n π \pm \frac{π}{6}, n \in Z

answer is C.

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Detailed Solution

2 \sin^{2} θ = \cos 2 θ

$1 - \cos 2 θ = \cos 2 θ$

$\cos 2 θ = \frac{1}{2} = \cos \frac{π}{3}$

$2 θ = 2 n π \pm \frac{π}{3}$ , $n \in Z$

$θ = n π \pm \frac{π}{6}$ , $n \in Z$

$\Rightarrow \frac{θ}{2} = \frac{n π}{2} \pm \frac{π}{12}, n \in Z$

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