Question

Solve : $2{\cos }^2\theta +\sin \theta \le 2$, where $\frac{\pi }{2}\le \theta \le \frac{3\pi }{2}$

• A. $\theta ϵ[\frac{\pi }{2},\frac{5\pi }{8}]\cup [\pi ,\frac{3\pi }{2}]$
• B. $\theta ϵ[\frac{\pi }{2},\frac{5\pi }{6}]\cup [\pi ,\frac{5\pi }{4}]$
• C. $\theta ϵ[\frac{\pi }{2},\frac{5\pi }{7}]\cup [\pi ,\frac{3\pi }{2}]$
• D. $\theta ϵ[\frac{\pi }{2},\frac{5\pi }{6}]\cup [\pi ,\frac{3\pi }{2}]$

Answer 1

The correct option is D $\theta ϵ[\frac{\pi }{2},\frac{5\pi }{6}]\cup [\pi ,\frac{3\pi }{2}]$

Given, $2{\cos }^2\theta +\sin \theta \le 2$
$\Rightarrow 2(1-{\sin }^2\theta )+\sin \theta \le 2$
$\Rightarrow -2{\sin }^2\theta +\sin \theta \le 0$
$\Rightarrow 2{\sin }^2\theta -\sin \theta \ge 0$
$\Rightarrow \sin \theta (2\sin \theta -1)\ge 0$
$\Rightarrow \sin \theta (\sin \theta -1/2)\ge 0$
which is possible, if $\sin \theta \le 0$ or $\sin \theta \ge \frac{1}{2}$
From the graph, we have
$\sin \theta \ge \frac{1}{2}$ $\Rightarrow$ $\frac{\pi }{2}\le \theta \le \frac{5\pi }{6}$
$\sin \theta \le 0$ $\Rightarrow$ $\pi \le \theta \le \frac{3\pi }{2}$
Hence, the required values of $\theta$ are given by,
$\theta ϵ[\frac{\pi }{2},\frac{5\pi }{6}]\cup [\pi ,\frac{3\pi }{2}]$