$\begin{array}{l}cos7x - \sqrt{3}sin3x = -\sqrt{2}\\ \Leftrightarrow \dfrac{1}{2}cos7x - \dfrac{\sqrt{3}}{2}sin7x = -\dfrac{\sqrt{2}}{2}\\ \Leftrightarrow sin\dfrac{\pi}{6}.cos7x - cos\dfrac{\pi}{6}.sin7x = sin(-\dfrac{\pi}{4})\\ \Leftrightarrow sin\left(\dfrac{\pi}{6} - 7x\right)=sin(-\dfrac{\pi}{4})\\ \Leftrightarrow \left[\begin{array}{l}\dfrac{\pi}{6} -7x = -\dfrac{\pi}{4} + k2\pi\\\dfrac{\pi}{6} - 7x=\dfrac{5\pi}{4} + k2\pi \end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}7x = \dfrac{5\pi}{12} + k2\pi\\ 7x=-\dfrac{13\pi}{12} + k2\pi \end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{5\pi}{84} + k\dfrac{2\pi}{7}\\ x=-\dfrac{13\pi}{84} + k\dfrac{2\pi}{7} \end{array}\right.\,\,\,\,(k \in \Bbb Z)\end{array}$
