If the sum of solutions of the system of equations $2{\sin }^2\mathrm{\theta }-\cos 2\mathrm{\theta }=0\text{ and }2{\cos }^2\mathrm{\theta }+3\sin \mathrm{\theta }=0$ interval $[0,2\mathrm{\pi }]\text{ is }\mathrm{k\pi }$, then k is equal to _______.

$\begin{array}{l}2{\sin }^2\mathrm{\theta }-\cos 2\mathrm{\theta }=0\\ 2{\sin }^2\mathrm{\theta }-\left(1-2{\sin }^2\mathrm{\theta }\right)=0\\ \Rightarrow {\sin }^2\mathrm{\theta }={\left(\frac{1}{2}\right)}^2\end{array}$

$\begin{array}{l}\mathrm{\theta }=\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6},\frac{11\mathrm{\pi }}{6}\\ 2{\cos }^2\mathrm{\theta }+3\sin \mathrm{\theta }=0\\ \Rightarrow 2{\sin }^2\mathrm{\theta }-3\sin \mathrm{\theta }-2=0\\ \therefore \sin \mathrm{\theta }=-\frac{1}{2}\\ \mathrm{\theta }=\frac{7\mathrm{\pi }}{6},\frac{11\mathrm{\pi }}{6}\end{array}$

So, the common solution is

$\begin{array}{l}\mathrm{\theta }=\frac{7\mathrm{\pi }}{6},\frac{11\mathrm{\pi }}{6}\\ \mathrm{Sum}=\frac{7\mathrm{\pi }+11\mathrm{\pi }}{6}=3\mathrm{\pi }=\mathrm{k\pi }\\ \mathrm{K}=3\end{array}$