If $\cos 3x+\sin \left(2x-\frac{7\pi }{6}\right)=-2$ then $x=$

We have

$\begin{array}{l}\cos 3x+\sin \left(2x-\frac{7\pi }{6}\right)=-2\\ \Rightarrow 1+\cos 3x+1+\sin \left(2x-\frac{7\pi }{6}\right)=0\\ \Rightarrow (1+\cos 3x)+1-\cos \left(2x-\frac{2\pi }{3}\right)=0\\ \Rightarrow 2{\cos }^2\frac{3x}{2}+2{\sin }^2\left(x-\frac{\pi }{3}\right)=0\\ \Rightarrow \cos \frac{3x}{2}=0\text{ and }\sin \left(x-\frac{\pi }{3}\right)=0\\ \Rightarrow \frac{3x}{2}=\frac{\pi }{2},\frac{3\pi }{2},\dots \text{ and }x-\frac{\pi }{3}=0,\pi ,2\pi \dots \\ \Rightarrow x=\frac{\pi }{3}\end{array}$

Therefore, the general solution is given by

$x=2k\pi +\frac{\pi }{3}=\frac{\pi }{3}(6k+1),\text{ where }k\in Z.$