Question

If $4^{2si{n}^{2}x}{.16}^{ta{n}^{2}x}{.2}^{4co{s}^{2}x}=256$ such that $0<x<\frac{\pi }{2}$ then $x$ is equal to ___________.

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{12}$
• D. $\frac{\pi }{24}$

Answer 1

The correct option is B $\frac{\pi }{4}$

$4^{2{\sin }^{2}x}.{16}^{{\tan }^{2}x}.2^{4{\cos }^{2}x}=256$

$\Rightarrow 2^{4{\sin }^{2}x}.2^{4{\tan }^{2}x}.2^{4{\cos }^{2}x}=2^8$

$\Rightarrow 2^{4{\sin }^{2}x+4{\tan }^{2}x+4{\cos }^{2}x}=2^8$

$\Rightarrow 4{\sin }^{2}x+4{\tan }^{2}x+4{\cos }^{2}x=8$

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

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

$\Rightarrow 1+{\tan }^{2}x=2$ since $({\sin }^{2}x+{\cos }^{2}x=1)$

$\Rightarrow {\sec }^{2}x=2$ since $1+{\tan }^{2}x={\sec }^{2}x$

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

$\Rightarrow \cos x=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \cos x=\frac{1}{\sqrt{2}}$ since $0<x<\frac{\pi }{2}$

$\Rightarrow x=\frac{\pi }{4}$