Question

Solve $2{\cos }^2\theta +\sin \theta \le 2$ where $\pi /2\le \theta \le 3\pi /2$.

Answer 1

$2{\cos }^2\theta +\sin \theta \le 2$
$2(1-{\sin }^2\theta )+\sin \theta \le 2$
$1-{\sin }^2\theta +\frac{1}{2}\sin \theta \le 1$
${\sin }^2\theta -\frac{1}{2}\sin \theta \ge 0$
$(\sin \theta -\frac{1}{4}{)}^2\ge \frac{1}{16}$
$-\frac{1}{4}\le \sin \theta -\frac{1}{4}\le \frac{1}{4}$
$0\le \sin \theta \le \frac{1}{2}$
Therefore ,$\theta \in [\frac{5\pi }{6},\pi ]$