Question

${\left(\sqrt{3}\sin x+\cos x\right)}^{\sqrt{\sqrt{3}sin2x-\cos 2x+2}}=4$

Answer 1

$\sqrt{\sqrt{3}.\sin 2x-\cos 2x+2}$

$\left(\sqrt{\sqrt{3}\sin x+\cos x}\right)$

$=2\{\frac{3}{2}\sin x+\frac{1}{2}\cos x\}\sqrt{2(\frac{\sqrt{3}}{2}\sin 2x-\frac{1}{2}\cos 2x)+2}$

$=4$

$=\sqrt{2(\sin 2x.\cos 30-\cos 2x.\sin 30)+2}$

$=2\{\sin x\cos 30+\cos x.\sin 30\}$

$=4$

$=2\left\{\sin \right(x+30\left)\right\}\sqrt{2\sin (2x-30)+2}$

$=2$

On comparing both side because of positive Natur of $\sin$

$2\sin (x+30)=2$

$\sin (x+30)=\sin {90}^o$

$x+30=90$

$x={60}^o$

$\sqrt{2\sin (2x-30)+2}=2$

$2\sin (2x-30)+2=4$

$2\sin (2x-30)=2$

$\sin (2x-30)=\sin 90$

$2x-30=90$

$2x=120$

$x={60}^o$