Q. The general value of $\theta$ satisfying $\sin^{2}{\theta + \sin\theta = 2}$ is
A) $n\pi + ( - 1)^{n}\frac{\pi}{6}$
B) $2n\pi + \frac{\pi}{4}$
C) $n\pi + ( - 1)^{n}\frac{\pi}{2}$
D) $n\pi + ( - 1)^{n}\frac{\pi}{3}$
Explanation
(c)
We have, $\sin^{2}{\theta + \sin\theta - 2} = 0$
$\Rightarrow (\sin{\theta - 1)(\sin{\theta + 2) = 0}}$
$\Rightarrow \sin{\theta = 1,\sin{\theta = - 2}}$
But $\sin{\theta \neq - 2}$
$\therefore\sin{\theta = 1 = \sin\frac{\pi}{2}}$
$\Rightarrow \ \theta = n\pi + ( - 1)^{n}\frac{\pi}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.