Giải phương trình $\sqrt 3 {\cos ^2}x + \sin 2x - \sqrt 3 {\sin ^2}x = 1$.

Ta có:

$\begin{array}{l}\,\,\,\,\,\sqrt 3 {\cos ^2}x + \sin 2x - \sqrt 3 {\sin ^2}x = 1\\ \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\left( {1 + \cos 2x} \right) + \sin 2x - \dfrac{{\sqrt 3 }}{2}\left( {1 - \cos 2x} \right) = 1\\ \Leftrightarrow \sqrt 3  + \sqrt 3 \cos 2x + 2\sin 2x - \sqrt 3  + \sqrt 3 \cos 2x = 2\\ \Leftrightarrow 2\sin 2x + 2\sqrt 3 \cos 2x = 2\\ \Leftrightarrow \sin 2x + \sqrt 3 \cos 2x = 1\\ \Leftrightarrow \dfrac{1}{2}\sin 2x + \dfrac{{\sqrt 3 }}{2}\cos 2x = \dfrac{1}{2}\\ \Leftrightarrow \sin 2x\cos \dfrac{\pi }{3} + \cos 2x\sin \dfrac{\pi }{3} = \dfrac{1}{2}\\ \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{3}} \right) = \sin \dfrac{\pi }{6}\\ \Leftrightarrow \left[ \begin{array}{l}2x + \dfrac{\pi }{3} = \dfrac{\pi }{6} + k2\pi \\2x + \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{\pi }{4} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}$

Vậy nghiệm của phương trình là $x =  - \dfrac{\pi }{{12}} + k\pi ,\,\,x = \dfrac{\pi }{4} + k\pi \,\,\left( {k \in \mathbb{Z}} \right)$.