$\begin{array}{l}
y = 2{\cos ^2}x - 2\sqrt 3 \sin x\cos x + 1\\
y = 2\left( {\dfrac{{1 + \cos 2x}}{2}} \right) - \sqrt 3 \sin 2x + 1\\
y = 1 + \cos 2x - \sqrt 3 \sin 2x + 1\\
y = \cos 2x - \sqrt 3 \sin 2x + 2 = 2\cos \left( {2x + \dfrac{\pi }{3}} \right) + 2\\
x \in \left[ {0;\dfrac{{7\pi }}{{12}}} \right] \Rightarrow 0 \le 2x \le \dfrac{{14\pi }}{{12}} = \dfrac{{7\pi }}{6} \Rightarrow \dfrac{\pi }{3} \le 2x + \dfrac{\pi }{3} \le \dfrac{{3\pi }}{2}\\
\Rightarrow \cos \left( {2x + \dfrac{\pi }{3}} \right) \in \left[ { - 1;\dfrac{1}{2}} \right]\\
\Rightarrow y \in \left[ {0;3} \right]\\
\Rightarrow \left\{ \begin{array}{l}
\max y = 3 \Rightarrow 2x + \dfrac{\pi }{3} = \pi + k2\pi \Rightarrow x = \dfrac{\pi }{3} + k\pi \\
\min y = 0 \Rightarrow 2x + \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \Rightarrow x = k\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$
