Q. If $5\cos{2\theta + 2\cos^{2}\frac{\theta}{2}} + 1 = 0,\ - \pi < \theta < \pi$, then $\theta$ is equal to
A) $\frac{\pi}{3}$
B) $\frac{\pi}{3},\cos^{- 1}\left( \frac{3}{5} \right)$
C) $\cos^{- 1}\left( \frac{3}{5} \right)$
D) $\frac{\pi}{3},\ \pi - \cos^{- 1}\left( \frac{3}{5} \right)$
Explanation
(d)
$5\cos{2\theta + 2\cos^{2}{\frac{\theta}{2} + 1 = 0}}$
$\Rightarrow \ \ 5\left( 2\cos^{2}{\theta - 1} \right) + \left( 1 + \cos\theta \right) + 1 = 0$
$\Rightarrow \ \ 10cos^{2}\theta + \cos{\theta - 3 = 0}$
$\Rightarrow \ \ \left( 5\cos{\theta + 3} \right)\left( 2\cos{\theta - 1} \right) = 0$
$\Rightarrow \ \cos{\theta = \frac{1}{2},\ \cos{\theta = - \frac{3}{5}\ \ \Rightarrow \ \ \theta = \frac{\pi}{3},\ \pi - \cos^{- 1}\left( \frac{3}{5} \right)}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.