$\begin{array}{l}
a){\cos ^2}x - {\sin ^2}x - 2\sqrt 3 \sin x\cos x = 1\\
 \Leftrightarrow \cos 2x - 2.\sin x\cos x\sqrt 3  = 1\\
 \Leftrightarrow \cos 2x - \sqrt 3 \sin 2x = 1\\
 \Leftrightarrow 2\cos \left( {2x + \dfrac{\pi }{6}} \right) = 1\\
 \Leftrightarrow \cos \left( {2x + \dfrac{\pi }{6}} \right) = \dfrac{1}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{6} = \dfrac{\pi }{3} + k2\pi \\
2x + \dfrac{\pi }{6} =  - \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{6} + k2\pi \\
2x =  - \dfrac{\pi }{2} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x =  - \dfrac{\pi }{4} + k\pi 
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
b)\sqrt 3 {\sin ^2}x + \left( {1 - \sqrt 3 } \right)\sin x\cos x - {\cos ^2}x + 1 - \sqrt 3  = 0\\
 \Leftrightarrow \sqrt 3 {\sin ^2}x + \sin x\cos x - \sqrt 3 \sin x\cos x - {\cos ^2}x + 1 - \sqrt 3  = 0\\
 \Leftrightarrow \sqrt 3 \left( {{{\sin }^2}x - \sin x\cos x - 1} \right) + \sin x\cos x - {\cos ^2}x + 1 = 0\\
 \Leftrightarrow \sqrt 3 \left( { - {{\cos }^2}x - \sin x\cos x} \right) + \sin x\cos x + {\sin ^2}x = 0\\
 \Leftrightarrow  - \cos x\sqrt 3 \left( {\cos x + \sin x} \right) + \sin x\left( {\sin x + \cos x} \right) = 0\\
 \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {\sin x - \sqrt 3 \cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x + \cos x = 0\\
\sin x - \sqrt 3 \cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
2\sin \left( {x - \dfrac{\pi }{3}} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = k\pi \\
x - \dfrac{\pi }{3} = k\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{3} + k\pi 
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$