Prove the following


Let $S=\left\{\theta \in[-2 \pi, 2 \pi]: 2 \cos ^{2} \theta+3 \sin \theta=0\right\}$. Then the sum of the elements of $S$ is:

  1. (1) $\frac{13 \pi}{6}$

  2. (2) $\frac{5 \pi}{3}$

  3. (3) $2 \pi$

  4. (4) $\pi$

Correct Option: , 3


$2 \cos ^{2} \theta+3 \sin \theta=0$

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

$\Rightarrow \sin \theta=-\frac{1}{2}$ or $\sin \theta=2 \rightarrow$ Not possibe

The required sum of all solutions in $[-2 \pi, 2 \pi]$ is

$=\left(\pi+\frac{\pi}{6}\right)+\left(2 \pi-\frac{\pi}{6}\right)+\left(-\frac{\pi}{6}\right)+\left(-\pi+\frac{\pi}{6}\right)=2 \pi$

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