${\cos }^{2}x-\sqrt{3}\sin 2x=1+{\sin }^{2}x$

$\iff {\cos }^{2}x-{\sin }^{2}x-\sqrt{3}\sin 2x=1\sqrt{b^2-4ac}$

$\iff \cos 2x-\sqrt{3}\sin 2x=1$

$\iff \frac{1}{2}\cos 2x-\frac{\sqrt{3}}{2}\sin 2x=\frac{1}{2}$

$\iff \cos 2x.\cos \frac{\pi }{3}-\sin 2x.\sin \frac{\pi }{3}=\frac{1}{2}$

$\iff \cos \left(2x+\frac{\pi }{3}\right)=\cos \frac{\pi }{3}$

$\iff \left[\begin{array}{l}2x+\frac{\pi }{3}=\frac{\pi }{3}+k2\pi \\ 2x+\frac{\pi }{3}=-\frac{\pi }{3}+k2\pi \end{array}\right.\iff \left[\begin{array}{l}x=k\pi \\ x=-\frac{\pi }{3}+k\pi \end{array}\right.\left(k\in \mathbb{Z}\right)$

Vậy tập họ nghiệm của phương trình là:  $S=\left\{k\pi ;-\frac{\pi }{3}+k\pi ,k\in \mathbb{Z}\right\}$.