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: (B), (D)

$\frac{\pi }{6}$
$\frac{5\pi }{6}$

Given, $\frac{{\sin }^{2}2x+4{\sin }^{4}x-4{\sin }^{2}x{\cos }^{2}x}{4-{\sin }^{2}2x-4{\sin }^{2}x}=\frac{1}{9}$

$\Rightarrow \frac{{\left(2\sin x\cos x\right)}^2+4{\sin }^{4}x-4{\sin }^{2}x{\cos }^{2}x}{4-{\left(2\sin x\cos x\right)}^2-4{\sin }^{2}x}=\frac{1}{9}\Rightarrow \frac{4{\sin }^{2}x{\cos }^{2}x+4{\sin }^{4}x-4{\sin }^{2}x{\cos }^{2}x}{4-4{\sin }^{2}x{\cos }^{2}x-4{\sin }^{2}x}=\frac{1}{9}\Rightarrow \frac{{\sin }^{4}x}{1-{\sin }^{2}x-{\sin }^{2}x{\cos }^{2}x}=\frac{1}{9}\Rightarrow \frac{{\sin }^{4}x}{{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x}=\frac{1}{9}\Rightarrow \frac{{\sin }^{4}x}{{\cos }^{2}x(1-{\sin }^{2}x)}=\frac{1}{9}\Rightarrow \frac{{\sin }^{4}x}{{\cos }^{4}x}=\frac{1}{9}\Rightarrow {\tan }^{4}x=\frac{1}{9}\Rightarrow \tan x=\pm \frac{1}{\sqrt{3}}\Rightarrow x=\frac{\pi }{6},\pi -\frac{\pi }{6}\Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6}$

So, options A and C are correct.