Question

The solution set for $2{\cos }^2\theta +\sin \theta \le 2,$where $\frac{\pi }{2}\le \theta \le \frac{3\pi }{2}$, is

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

Answer 1

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

​​​​$2{\cos }^2\theta +\sin \theta \le 2\Rightarrow 2(1-{\sin }^2\theta )+\sin \theta \le 2\Rightarrow -2{\sin }^2\theta +\sin \theta \le 0\Rightarrow 2{\sin }^2\theta -\sin \theta \ge 0\Rightarrow \sin \theta (2\sin \theta -1)\ge 0\Rightarrow \sin \theta (\sin \theta -\frac{1}{2})\ge 0\Rightarrow \sin \theta \le 0\text{or}\sin \theta \ge \frac{1}{2}$
Now,
$\sin \theta \ge \frac{1}{2}\Rightarrow \frac{\pi }{2}\le \theta \le \frac{5\pi }{6}\sin \theta \le 0\Rightarrow \pi \le \theta \le \frac{3\pi }{2}$

Hence, the required value of $\theta$ is given by
$[\frac{\pi }{2},\frac{5\pi }{6}]\cup [\pi ,\frac{3\pi }{2}]$