Question

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

• A. $\frac{13 \pi}{6}$
• B. $\pi$
• C. $2 \pi$
• D. $\frac{5 \pi}{3}$

Answer 1

Answer: (C)

$2 (1- sin^2 \theta) + 3sin \theta = 0$
$\Rightarrow \, \, 2 sin^2 \theta - 3 sin \theta - 2 = 0$
$\Rightarrow (2 sin \theta + 1 ) (sin \theta - 2) = 0$
$\Rightarrow \, \, sin \theta = - \frac{1}{2} ; sin \theta = 2 (reject)$
roots : $\pi + \frac{\pi}{6}, 2\pi - \frac{\pi}{6}, - \frac{\pi}{6}, - \pi + \frac{\pi}{6}$
$\Rightarrow $ sum of values = $2\pi$