Given,

$2{\sin }^2\theta -\cos 2\theta =0$

$\Rightarrow 2{\sin }^2\theta -\left(1-2{\sin }^2\theta \right)=0$

$\Rightarrow {\sin }^2\theta ={\left(\frac{1}{2}\right)}^2$

$\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$

Also given, $2{\cos }^2\theta +3\sin \theta =0$

$\Rightarrow 2{\sin }^2\theta -3\sin \theta -2=0$

$\Rightarrow \sin \theta =-\frac{1}{2}$

$\Rightarrow \theta =\frac{7\pi }{6},\frac{11\pi }{6}$

So, the common solution is

$\theta =\frac{7\pi }{6},\frac{11\pi }{6}$

Sum $=\frac{7\pi +11\pi }{6}=3\pi =\mathrm{k}\pi$

$\Rightarrow K=3$