If 2 sin2π2cos2x=1-cosπ sin2x, x≠2n+1π2, n∈I then cos2x is equal to

2sin2π2cos2x=2sin2π2sin2x⇒cos2x=sin 2x⇒cos xcos x-2 sinx=0cos x≠0x≠2n+1n2,tan x=12⇒cos 2x=1-tan2x1+tan2x=35

If 2 sin2π2cos2x=1-cosπ sin2x, x≠2n+1π2, n∈I then cos2x is equal to

2sin2π2cos2x=2sin2π2sin2x⇒cos2x=sin 2x⇒cos xcos x-2 sinx=0cos x≠0x≠2n+1n2,tan x=12⇒cos 2x=1-tan2x1+tan2x=35

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If $2 \sin^{2} (\frac{π}{2} \cos^{2} x) = 1 - \cos (π \sin 2 x), x \neq (2 n + 1) \frac{π}{2}, n \in I$ then cos2x is equal to

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$\frac{3}{5}$

$\frac{1}{5}$

$\frac{4}{5}$

answer is B.

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Detailed Solution

$2 \sin^{2} (\frac{π}{2} \cos^{2} x) = 2 \sin^{2} (\frac{π}{2} \sin 2 x) \Rightarrow \cos^{2} x = \sin 2 x$

$\Rightarrow \cos x (\cos x - 2 sinx) = 0 \cos x \neq 0 x \neq (2 n + 1) \frac{n}{2}, \tan x = \frac{1}{2}$ $\Rightarrow \cos 2 x = \frac{1 - \tan^{2} x}{1 + \tan^{2} x} = \frac{3}{5}$

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If 2 sin2π2cos2x=1-cosπ sin2x, x≠2n+1π2, n∈I then cos2x is equal to

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If $2 \sin^{2} (\frac{π}{2} \cos^{2} x) = 1 - \cos (π \sin 2 x), x \neq (2 n + 1) \frac{π}{2}, n \in I$ then cos2x is equal to