Question

$3^{\sin 2x+\cos 2x+1}+3^{2-\sin 2x-\cos 2x}=28$ is satisfied by-

• A. Those values of for which $\tan x=-1$
• B. Those values of x for which $\tan x=-\frac{1}{2}$
• C. Those values of x for which $\cos x=0$
• D. Those values of x for which $\tan x=1$

Answer 1

Answer: (A), (C)

The correct options are
A Those values of for which $\tan x=-1$
C Those values of x for which $\cos x=0$

Given, $3^{\sin 2x+\cos 2x+1}+3^{2-\sin 2x-\cos 2x}=28$
Taking $3^{\sin 2x+\cos 2x+1}=y$,
$y+\frac{27}{y}=28$ or $y^2-28y+27=0$ or $y=1,27=3^o,3^3$
$\therefore \sin 2x+\cos 2x+1=0,3$
But the maximum value of $\sin 2x+\cos 2x=\sqrt{2}$.
$\therefore$ we get only $\sin 2x+\cos 2x+1=0$
$\Rightarrow \frac{2\tan x}{1+{\tan }^{2}x}+\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}+1=0\Rightarrow \tan x=-1$
Also, $\sin 2x+\cos 2x+1=0$
$\Rightarrow 2\sin x\cos x+{\cos }^{2}x=0\Rightarrow \cos x=0$ or $2\sin x+\cos x=0$
$\cos x=0$ satisfies the equation, but $\tan x=-\frac{1}{2}$ does not exist.