Question

If $\frac{{\sin }^{2}2x+4{\sin }^{4}x-4{\sin }^{2}x{\cos }^{2}x}{4-{\sin }^{2}2x-4{\sin }^{2}x}=\frac{1}{9}$ and $0<x<\pi$, then the value of $x$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{2\pi }{3}$
• D. $\frac{5\pi }{6}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{\pi }{3}$
C $\frac{2\pi }{3}$

Num=${\sin }^{2}2x+4{\sin }^{4}x-4{\sin }^{2}x.{\cos }^{2}x$
$=4{\sin }^{2}x.{\cos }^{2}x-4{\sin }^{4}x-4{\sin }^{2}x.{\cos }^{2}x$
$=4{\sin }^{4}x$ ...(i)
$4-{\sin }^{2}2x-4{\sin }^{2}x$
Den=$=4-4{\sin }^{2}x.{\cos }^{2}x-4{\sin }^{2}x$
$=4({\sin }^{2}x+{\cos }^{2}x)-4{\sin }^{2}x.{\cos }^{2}x-4{\sin }^{2}x$
$=4{\cos }^{2}x-4{\sin }^{2}x.{\cos }^{2}x$
$=4{\cos }^{2}x(1-{\sin }^{2}x)$
$=4{\cos }^{4}x$ ...(ii)
Hence $\frac{i}{ii}$ gives
${\tan }^{4}x=\frac{1}{9}$
${\tan }^{2}x=\frac{\pm 1}{3}$
${\tan }^{2}x=\frac{-1}{3}$ is not possible.
Hence
$\tan x=\frac{\pm 1}{\sqrt{3}}$
$x=\frac{\pi }{3}$ and $x=\frac{-\pi }{3},\frac{2\pi }{3}$.