Q. If $2\sin^{2}{\theta = 3\cos{\theta,\ 0 \leq \theta \leq 2\pi,}}$ then $\theta$ is equal to
A) $\frac{\pi}{6},\frac{5\pi}{6}$
B) $\frac{\pi}{3},\frac{2\pi}{3}$
C) $\frac{\pi}{3},\frac{5\pi}{3}$
D) $\frac{\pi}{2},\ \pi$
Explanation
(c)
Since, $2\sin^{2}{\theta = 3\cos\theta}$
$\Rightarrow \ 2 - 2\cos^{2}{\theta = 3\cos\theta}$
$\Rightarrow \ 2\cos^{2}{\theta + 3\cos{\theta - 2 = 0}}$
$\Rightarrow \ (2\cos{\theta - 1)(\cos{\theta + 2) = 0}}$
$\Rightarrow \ \cos\theta = \frac{1}{2}\ \ \ (\because\cos{\theta \neq - 2)}$
$\therefore\ \ \theta = \frac{\pi}{3},\frac{5\pi}{3}\ \ \ \ (\because 0 \leq \theta \leq 2\pi)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.