Question

If $2{\sin }^2\theta -5\sin \theta +2>0,\theta \in (0,2\pi )$, then $\theta \in$.

• A. $(\frac{5\pi }{6},2\pi )$
• B. $(0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$
• C. $(0,\frac{\pi }{6})$
• D. $(\frac{\pi }{80},\frac{\pi }{6})$

Answer 1

Answer: (B)

$(0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$

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

$\therefore \sin \theta <\frac{1}{2}=\sin \frac{\pi }{6}=\sin (\pi -\frac{\pi }{6})=\sin \frac{5\pi }{6}$

$\therefore \theta \in (0,\frac{\pi }{6})$ or $\theta \in (\frac{5\pi }{6},2\pi )$

$\therefore \theta \in (0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$.