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Q.

The general solution of $2^{\cos 2 x} + 1 = {3.2}^{- \sin^{2} x}$

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a

$n π, n \in I$

b

$n π \pm \frac{π}{2}, n \in I$

c

$n π - π, n \in I$

d

$n π + π, n \in I$

answer is A, B.

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Detailed Solution

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$2^{1 - 2 \sin^{2} x} + 1 = 3 (2^{- \sin^{2} x}) let 2^{- \sin^{2} x} = t$

$\Rightarrow 2 t^{2} + 1 = 3 t \Rightarrow t = 1 / 2 and 1$

$t = 1 / 2 \Rightarrow x = n π \pm π / 2, t = 1 \Rightarrow x = n π$

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