The sum of all values of $\theta \in \left(0,\frac{\pi }{2}\right)$ satisfying  ${\sin }^{2}2\theta +{\cos }^{4}2\theta =\frac{3}{4}$, is 

Solution:

we have , 

${\sin }^{2}2\theta +{\cos }^{4}2\theta =\frac{3}{4}$ 

$\begin{array}{l}\Rightarrow 1-{\cos }^{2}2\theta +{\cos }^{4}2\theta =\frac{3}{4}\\ \Rightarrow 4{\cos }^{4}2\theta -4{\cos }^{2}2\theta +1=0\\ \Rightarrow {\left(2{\cos }^{2}2\theta -1\right)}^2=0\Rightarrow \cos 4\theta =0\Rightarrow \theta =\frac{\pi }{8},\frac{3\pi }{8}\end{array}$

Hence , required sum $=\frac{\pi }{8}+\frac{3\pi }{8}=\frac{\pi }{2}$