Let S = \ θ ∈ [- 2π, 2π]: 2 ^ 2 θ + 3sinθ = 0\ . Then the sum of...

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

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

$⇒ 2\sin^{2}\theta - 3sin\theta - 2 = 0$

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

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

$⇒ sin\theta = - \dfrac{1}{2}$

Hence, solutions in \left[- 2\pi, 2\pi\right]$are<br />$- \pi + \dfrac{\pi}{6}, - \dfrac{\pi}{6}, \pi + \dfrac{\pi}{6}, 2\pi - \dfrac{\pi}{6}$</p><p>Sum of the elements is$= 2\pi .