Question

The least positive value of $x$ satisfying $\frac{{\sin }^{2}2x+4{\sin }^{4}x-4{\sin }^{2}x{\cos }^{2}x}{4-{\sin }^{2}2x-4{\sin }^{2}x}=\frac{1}{9}$ is

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

Answer 1

The correct option is B $\frac{\pi }{6}$

We have,

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

$\Rightarrow \frac{{\sin }^{2}2x+4{\sin }^{4}x-{\left(2\sin x\cos x\right)}^2}{4-{\left(2\sin x\cos x\right)}^2-4{\sin }^{2}x}=\frac{1}{9}$

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

$\Rightarrow \frac{4{\sin }^{4}x}{4(1-{\sin }^{2}x{\cos }^{2}x-{\sin }^{2}x)}=\frac{1}{9}$

$\Rightarrow \frac{{\sin }^{4}x}{1-{\sin }^{2}x-{\sin }^{2}x{\cos }^{2}x}=\frac{1}{9}$

$\Rightarrow \frac{{\sin }^{4}x}{{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x}=\frac{1}{9}$

$\Rightarrow \frac{{\sin }^{4}x}{{\cos }^{2}x(1-{\sin }^{2}x)}=\frac{1}{9}$

$\Rightarrow \frac{{\sin }^{4}x}{{\cos }^{2}x{\cos }^{2}x}=\frac{1}{9}$

$\Rightarrow \frac{{\sin }^{4}x}{{\cos }^{4}x}=\frac{1}{9}$

$\Rightarrow {\tan }^{4}x=\frac{1}{9}$

$\Rightarrow {\tan }^{4}x={\left(\frac{1}{\sqrt{3}}\right)}^4$

On comparing both side and we get,

$\mathrm{tanx}=\frac{1}{\sqrt{3}}$

$\Rightarrow \tan x=\tan \frac{\pi }{6}$

On comparing that,

$x=\frac{\pi }{6}$

Hence, this is the answer.