Question

If $0<\Theta <2\pi$, then the intervals of values of $\Theta$ for which 2 ${\sin }^2\Theta -5\sin \Theta +2>0$, is

• A. $(0,\frac{\pi }{6}\left)u\right(\frac{5\pi }{6},2\pi )$
• B. $(\frac{\pi }{8}\frac{5\pi }{6})$
• C. $(0,\frac{\pi }{8}\left)u\right(\frac{\pi }{6}\frac{5\pi }{6}1$
• D. $(\frac{41\pi }{48},\pi )$

Answer 1

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

Given, $2{\sin }^2\Theta -5\sin \Theta +2>0$

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

$\Rightarrow \sin \Theta <\frac{1}{2}$

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