The number of solutions of the equation $2 {\sin }^{2}x+\sqrt{3} \cos x 1= 0$ in $[-\mathrm{\pi },3\mathrm{\pi }]$ is

$2{\sin }^{2}x+\sqrt{3}\cos x+1=0$

$$\begin{gathered} \Rightarrow 2(1-{\cos }^{2}x)+\sqrt{3} \cos x+1=0 \\ \Rightarrow 2{\cos }^{2}x-\sqrt{3} \cos x-3=0 \end{gathered}$$

Solving, we get$\cos x=-\frac{\sqrt{3}}{2},\sqrt{3}$but $\cos x\ne \sqrt{3}$

$$\begin{gathered} \therefore \cos x=-\frac{\sqrt{3}}{2}=\cos \frac{5\mathrm{\pi }}{2} \therefore x=2n\mathrm{\pi }\pm \frac{5\mathrm{\pi }}{6} \\ \Rightarrow \mathrm{x}=\pm \frac{5\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6},\frac{17\mathrm{\pi }}{6}, \end{gathered}$$