sin 2 2$\theta$ + cos 4 2$\theta$ = {3 \over 4}, $$$$\theta $\in$ $\left( {0,{\pi \over 2}} \right)$ $\Rightarrow$ 1 $-$ cos 2 2$\theta$ + cos 4 2$\theta$ = ${3 \over 4}$ $\Rightarrow$ 4cos2$\theta$ $-$ 4cos 2 2$\theta$ + 1 = 0 $\Rightarrow$ (2cos 2 2$\theta$ $-$ 1) 2 = 0 $\Rightarrow$ cos 2 2$\theta$ = ${1 \over 2}$ = cos 2 ${{\pi \over 4}}$ $\Rightarrow$ 2$\theta$ = n$\pi$ $\pm$ ${\pi \over 4}$, n $\in$ I $\Rightarrow$ $\theta$ = ${{n\pi } \over 2} \pm {\pi \over 8}$ $\Rightarrow$ $\theta$ = ${\pi \over 8},{\pi \over 2} - {\pi \over 8}$ Sum of solutions ${\pi \over 2}$