Question

The most general solutions of $2^{\sin x}+2^{\cos x}=2^{1-\frac{1}{\sqrt{2}}}$ are

• A. $n\pi -\frac{\pi }{4}$
• B. $n\pi +\frac{\pi }{4}$
• C. $n\pi +{(-1)}^n\frac{\pi }{4}$
• D. $2n\pi \pm \frac{\pi }{4}$

Answer 1

The correct option is B $n\pi +\frac{\pi }{4}$

$2^{\sin x}+2^{\cos x}=2^{1-\frac{1}{\sqrt{2}}}$
Since, $AM\ge GM$
$\frac{2^{\sin x}+2^{\cos x}}{2}\ge \sqrt{2^{\sin x+\cos x}}$
$\Rightarrow 2^{\sin x}+2^{\cos x}\ge 2\sqrt{{(2}^{\sin x+\cos x})}$ ....(1)
We know that $-\sqrt{1+1}\le \sin x+\cos x\le \sqrt{1+1}$
$\Rightarrow -\sqrt{2}\le \sin x+\cos x\le \sqrt{2}$
So, minimum value of $\sin x+\cos x$ is $-\sqrt{2}$
So, by (1),
$2^{\sin x}+2^{\cos x}\ge 2(\sqrt{2^{-\sqrt{2}}})=2^{1-1/\sqrt{2}}$
The equation holds when equality holds
$2^{\sin x}=2^{\cos x}$
$\Rightarrow \sin x=\cos x$
$\Rightarrow \tan x=1$
$\Rightarrow x=n\pi +\frac{\pi }{4}$