Question

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

• A. those values of $x$ for which $\tan x=-1$
• B. those values of $x$ for which $\cos x=-\frac{1}{2}$
• 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)

those values of $x$ for which $\cos x=0$
those values of $x$ for which $\tan x=-1$

$3^{(\sin 2x+2{\cos }^{2}x)}+3^{(1-\sin 2x+2{\sin }^{2}x)}=28$

$3^{(\sin 2x+2{\cos }^{2}x)}+3^{(1-\sin 2x+2-2{\cos }^{2}x)}=28$

$3^{(\sin 2x+2{\cos }^{2}x)}+3^{(3-(\sin 2x+2{\cos }^{2}x\left)\right)}=28$

Put $3^{(\sin 2x+2{\cos }^{2}x)}=y$

$y+\frac{27}{y}=28$

$\Rightarrow y^2-28y+27=0$
$(y-27)(y-1)=0$
$\Rightarrow y=27,1$
If $y=27$

$3^{(\sin 2x+2{\cos }^{2}x)}=3^3$

$\Rightarrow$ $\sin 2x+2{\cos }^{2}x=3$

$\sin 2x+\cos 2x=2$
$\Rightarrow \sin (2x+\frac{\pi }{4})=\sqrt{2}$ (not possible)
If $y=1$
$3^{(\sin 2x+2{\cos }^{2}x)}=3^0$

$\sin 2x+2{\cos }^{2}x=0$
$2\cos x(\sin x+\cos x)=0$

Either $\cos x=0$
or $\sin x+\cos x=0$$⟹\tan x=-1$