Question

If $\frac{si{n}^{2}2x+4si{n}^{4}x-4si{n}^{2}xco{s}^{2}x}{4-si{n}^{2}2x-4si{n}^{2}x}=\frac{1}{9}$ and 0 < x < $\pi$, then value of x is-

• A. $\pi /3$
• B. $\pi /6$
• C. $2\pi /3$
• D. $\pi$

Answer 1

The correct option is B $\pi /6$

$\frac{{\sin }^{2}2x+4{\sin }^{4}x-4{\sin }^{2}x{\cos }^{2}x}{4-{\sin }^{2}2x-4{\sin }^{2}x}=\frac{1}{9}$
$\frac{4{\sin }^{2}x{\cos }^{2}x+4{\sin }^{4}x-4{\sin }^{2}x{\cos }^{2}x}{4-4{\sin }^{2}x{\cos }^{2}x-4{\sin }^{2}x}=\frac{1}{9}$
$\frac{4{\sin }^{4}x}{4-4{\sin }^{2}x{\cos }^{2}x-4{\sin }^{2}x}=\frac{1}{9}$
$\Rightarrow 32{\sin }^{4}x+8{\sin }^{2}x-4=0$
Let ${\sin }^{2}x=t$
$\Rightarrow 32t^2+8t-4=0$
$\Rightarrow t=-\frac{1}{2},t=\frac{1}{4}$
$\Rightarrow {\sin }^{2}x=-\frac{1}{2}$ (not possible),${\sin }^{2}x=\frac{1}{4}$
$\Rightarrow \sin x=\frac{1}{2}$ for $ 0 < x < \pi $
$\sin x=\sin \frac{\pi }{6}$
$\Rightarrow x=\frac{\pi }{6}$ for $0<x<\pi$