Question

For $xϵ(-\pi ,\pi )$ find the value of $x$ for which the given equation. $(\sqrt{3}\sin x+\cos x{)}^{\sqrt{\sqrt{3}\sin 2x-\cos 2x+2}}=4$ is satisfied.

• A. $x=\frac{\pi }{3}$
• B. $x=-\frac{\pi }{3}$
• C. $x=\frac{2\pi }{3}$
• D. $x=\frac{\pi }{6}$

Answer 1

The correct option is A $x=\frac{\pi }{3}$

Multiplying and dividing by $2$, we get
$2\left(\sin \right(x+\frac{\pi }{6}){)}^{\sqrt{2-2\cos (2x+\frac{\pi }{3})}}=2^2$
$2\left(\sin \right(x+\frac{\pi }{6}){)}^{\sqrt{4{\sin }^2(x+\frac{\pi }{6})}}=2^2$
$2\sin (x+\frac{\pi }{6}{)}^{2\sin (x+\frac{\pi }{6})}=2^2$
On comparing powers, we get
$\sin (x+\frac{\pi }{6})=1$
$x+\frac{\pi }{6}=\frac{\pi }{2}$
$x=\frac{\pi }{3}$