Sum of the root of the equation $2{\sin }^2\theta +{\sin }^{2}2\theta =2;0\le \theta \le \pi /2$ is

$4{\sin }^2\theta {\cos }^2\theta =2\left(1-{\sin }^2\theta \right)$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(2{\sin }^2\theta -1\right){\cos }^2\theta \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\end{array}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\sin }^2\theta \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=1/2\end{array}$or $c{\mathrm{os}}^2\theta \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0$

$\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\theta \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\pi /4$ or $\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\theta \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\pi /2$