2cos 2 $\theta$ + 3sin$\theta$ = 0 $\Rightarrow$ 2(1 - sin 2 $\theta$) + 3sin$\theta$ = 0 $\Rightarrow$ 2sin 2 $\theta$ - 3sin$\theta$ - 2 = 0 $\Rightarrow$ 2sin 2 $\theta$ - 4sin$\theta$ + sin$\theta$ - 2 = 0 $\Rightarrow$ (2sin$\theta$ + 1)(sin$\theta$ - 2) = 0 $\therefore$ sin$\theta$ = 2 (Not possible) sin$\theta$ = $- {1 \over 2}$ $\therefore$ $\theta$ = n$\pi$ + (-1) n $\left( { - {\pi \over 6}} \right)$ When n = 0, $\theta$ = ${ - {\pi \over 6}}$ When n = 1, $\theta$ = ${{7\pi } \over 6}$ When n = -1, $\theta$ = ${-{5\pi } \over 6}$ When n = 2, $\theta$ = ${{11\pi } \over 6}$ $\therefore$ Sum of all solutions in [–2$\pi$, 2$\pi$] is = $- {\pi \over 6} + {{7\pi } \over 6} - {{5\pi } \over 6} + {{11\pi } \over 6}$ = 2$\pi$