Question

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

• A. $(0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$
• B. $(\frac{\pi }{8},\frac{\pi }{6})$
• C. $(0,\frac{\pi }{8})\cup (\frac{\pi }{6},\frac{5\pi }{6})$
• D. $(0,\frac{\pi }{3})\cup (\frac{5\pi }{6},2\pi )$

Answer 1

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

$2{\sin }^2\theta -5\sin \theta +2>0$
$\Rightarrow (\sin \theta -2)(2\sin \theta -1)>0$
$\Rightarrow 2\sin \theta -1<0[\because (\sin \theta -2)<0]$
$\Rightarrow \sin \theta <\frac{1}{2}$

When $\sin x=\frac{1}{2},$ we have
$x=\frac{\pi }{6},\frac{5\pi }{6}$ for $x\in (0,2\pi )$


From the graph, it is clear that $\sin x<\frac{1}{2}$ when $x\in (0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$