Question

If $\theta \in (0,2\pi )$ then the interval of values of $\theta$ for which $2{\sin }^2\theta -5\sin \theta +2>0$ is

• A. $(\frac{\pi }{8},\frac{5\pi }{6})$
• B. $(\frac{41\pi }{48},\pi )$
• C. $(0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$
• D. $(0,\frac{\pi }{8})\cup (\frac{\pi }{6},\frac{5\pi }{6})$

Answer 1

The correct option is C $(0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$

$2{\sin }^2\theta -5\sin \theta +2>0$

$\Rightarrow (2\sin \theta -1)(\sin \theta -2)>0$

But we know, $-1\le \sin \theta \le 1\Rightarrow \sin \theta -2<0$

$\Rightarrow 2\sin \theta -1<0\Rightarrow -1\le \sin \theta <\frac{1}{2}$

Hence solution in given interval is,
$\theta \in (0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$