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

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$n π + {(- 1)}^{n} \frac{π}{4}$

$2 n π \pm \frac{π}{4}$

$n π - \frac{π}{4}$

$n π + \frac{π}{4}$

answer is B.

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Detailed Solution

$\frac{2^{\sin x} + 2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}}$

$2^{\sin x} + 2^{\cos x} \geq 2 \cdot \sqrt{2^{\sin x + \cos x}}$

But minimum value of $\sin x + \cos x$ is $- \sqrt{2}$

$∴ 2^{\sin x} + 2^{\cos x} \geq 2 \cdot \sqrt{2^{- \sqrt{2}}} = 2^{1 - \frac{1}{\sqrt{2}}}$

The equation holds when $2^{\sin x} = 2^{\cos x}$

$\Rightarrow \sin x = \cos x$ or $\tan x = 1$

$\Rightarrow x = n π + \frac{π}{4}$ .

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