For $n\in Z,2{\sin }^2\theta =\cos 2\theta$  implies $\frac{\theta }{2}=$

$1-\cos 2\theta =\cos 2\theta$

$\cos 2\theta =\frac{1}{2}=\cos \frac{\pi }{3}$

$2\theta =2n\pi \pm \frac{\pi }{3}$ , $n\in Z$

$\theta =n\pi \pm \frac{\pi }{6}$ , $n\in Z$

$\Rightarrow \frac{\theta }{2}=\frac{n\pi }{2}\pm \frac{\pi }{12},n\in Z$