Q. The general value of $\theta$ satisfying the equation $2\sin^{2}{\theta - 3\sin{\theta - 2 = 0}}$ is
A) $n\pi + ( - 1)^{n + 1}\frac{\pi}{6}$
B) $n\pi + ( - 1)^{n}\frac{\pi}{2}$
C) $n\pi + ( - 1)^{n}\frac{5\pi}{6}$
D) $n\pi + ( - 1)^{n}\frac{7\pi}{6}$
Explanation
(a)
$2\sin^{2}{\theta - 3\sin{\theta - 2 = 0}}$
$\Rightarrow \ \ \left( 2\sin{\theta + 1} \right)\left( \sin{\theta - 2} \right) = 0$
$\Rightarrow \ \ \sin{\theta = - \frac{1}{2}}\ \ \ \ \ \ \ \ \lbrack\because\sin{\theta \neq 2}\rbrack$
$\Rightarrow \ \sin{\theta = \sin\left( - \frac{\pi}{6} \right)}$
$\Rightarrow \ \ \ \ \theta = n\pi + ( - 1)^{n}\left( - \frac{\pi}{6} \right)$
$\Rightarrow \ \ \theta = n\pi + ( - 1)^{n + 1}\ \frac{\pi}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.