Let   S = θ ∈ - 2 π , 2 π : 2 cos 2 θ + 3 sin θ = 0 . Then the sum of…

The given equation $2{\cos }^2\theta +3\sin \theta =0$ can be written as

$2\left(1-{\sin }^2\theta \right)+3\sin \theta =0$

$\Rightarrow 2{\mathrm{s}\mathrm{i}\mathrm{n}}^2\theta -3\mathrm{s}\mathrm{i}\mathrm{n}\theta -2=0$

$\Rightarrow \left(2\sin \theta +1\right)\left(\sin \theta -2\right)=0$

$\Rightarrow \sin \theta =2, -\frac{1}{2}, \left(\sin \theta \ne 2\right)$

$\Rightarrow \sin \theta =-\frac{1}{2}$

Hence, solutions in $\left[-2\pi , 2\pi \right]$ are
$-\pi +\frac{\pi }{6}, -\frac{\pi }{6}, \pi +\frac{\pi }{6}, 2\pi -\frac{\pi }{6}$

Sum of the elements is $=2\pi .$