Answer
$\sqrt 3 \sin 2x+ \cos 2x=2 \sin 2 (x+\dfrac{\pi}{12})$
It has maximum value =2 and phase shift $=\dfrac{\pi}{12}$
Work Step by Step
Here, we have $\sqrt 3 \sin 2x+ \cos 2x=R \sin (2x-\phi)$
Since, the angle $\phi$ lies in the fourth quadrant so, $\phi=-\dfrac{\pi}{6}$
Thus, $\cos \phi=\cos \dfrac{\pi}{6}=\dfrac{\sqrt 3}{2}$; $\sin \phi=\sin \dfrac{\pi}{6}=\dfrac{-1}{2}$
Thus, $R^2(\cos^2 \phi+\sin^2 \phi)=(\sqrt 3)^2+(-1)^2$
This gives: $R=2$
since, the trigonometric function has period of $2 \pi$, so we will have to add $2 \pi$
$\sqrt 3 \sin 2x+ \cos 2x=R \sin (2x-\phi)$
Thus, we have $\sqrt 3 \sin 2x+ \cos 2x=2 \sin 2 (x+\dfrac{\pi}{12})$
It has maximum value =2 and phase shift $=\dfrac{\pi}{12}$
