The only value of x for which 2^{sin x} + 2^{ | vedclass

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(A) By the $A.M.-G.M.$ inequality,we have $\frac{2^{\sin x} + 2^{\cos x}}{2} \ge \sqrt{2^{\sin x} \cdot 2^{\cos x}}$.This simplifies to $2^{\sin x} +

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$\frac{5\pi}{4}$

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(A) By the $A.M.-G.M.$ inequality,we have $\frac{2^{\sin x} + 2^{\cos x}}{2} \ge \sqrt{2^{\sin x} \cdot 2^{\cos x}}$.This simplifies to $2^{\sin x} + 2^{\cos x} \ge 2 \cdot 2^{\frac{\sin x + \cos x}{2}} = 2^{1 + \frac{\sin x + \cos x}{2}}$.We know that $\sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})$,which has a minimum value of $-\sqrt{2}$.Thus,the minimum value of the expression $2^{\sin x} + 2^{\cos x}$ is $2^{1 - \frac{\sqrt{2}}{2}} = 2^{1 - (1/\sqrt{2})}$.The inequality $2^{\sin x} + 2^{\cos x} > 2^{1 - (1/\sqrt{2})}$ holds for all $x$ except where $\sin x + \cos x = -\sqrt{2}$.$\sin x + \cos x = -\sqrt{2}$ occurs when $\sin(x + \frac{\pi}{4}) = -1$,which implies $x + \frac{\pi}{4} = \frac{3\pi}{2}$,so $x = \frac{5\pi}{4}$.

The only value of $x$ for which $2^{\sin x} + 2^{\cos x} > 2^{1 - (1/\sqrt{2})}$ holds,is

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