Q. The set of values of $\theta$ satisfying the inequation $2\sin^{2}{\theta - 5\sin{\theta + 2 > 0}}$, where $0 < \theta < 2\pi$, is
A) $\left( 0,\ \frac{\pi}{6} \right) \cup \left( \frac{5\pi}{6},\ 2\pi \right)$
B) $\left\lbrack 0,\ \frac{\pi}{6} \right\rbrack \cup \left\lbrack \frac{5\pi}{6},\ 2\pi \right\rbrack$
C) $\left\lbrack 0,\ \frac{\pi}{3} \right\rbrack \cup \left\lbrack \frac{2\pi}{3},\ 2\pi \right\rbrack$
D) None of these
Explanation
(a)
Given, $2\sin^{2}{\theta - 5\sin{\theta + 2 > 0}}$
$\Rightarrow \ \ \left( 2\sin{\theta - 1} \right)\left( \sin{\theta - 2} \right) > 0$
[where, $\left( \sin{\theta - 2} \right) < 0$ for all $\theta \in R$]
[image:3_image10.emf]
$\left( 2\sin{\theta - 1} \right) < 0\ \ \ \Rightarrow \ \sin{\theta < \frac{1}{2}}$
It is clear from the figure
$\theta \in \left( 0,\frac{\pi}{6} \right) \cup \left( \frac{5\pi}{6},\ 2\pi \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.