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If 1 + sinx sin 2x = cosx cos 2x, then x =

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$2 n π, n \in Z$

$\frac{2 n π}{3}, n \in Z$

$\frac{2 n π}{5}, n \in Z$

$\frac{2 n π}{7}, n \in Z$

answer is B.

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Detailed Solution

$1 + \sin x \sin 2 x = \cos x \cos 2 x \cos 2 x \cos x - \sin 2 x \sin x = 1 \cos 3 x = 1 = \cos 2 nπ x = \frac{2 nπ}{3}$

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