Question

The solution set of inequality $\left(2{\sin }^{2}x-5\sin x+2\right)>0$ if $x\in [0,2\pi ]$ is

• A. $[0,\frac{\pi }{6}]$
• B. $[0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi ]$
• C. $(0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$
• D. none of these

Answer 1

Answer: (C)

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

$2{\sin }^{2}x-5\sin x+2>0\Rightarrow (2\sin x-1)(\sin x-2)>0$

Here, it is clear that $(\sin x-2)$ is negative as $\sin x<2$

Hence, $(2\sin x-1)<0\Rightarrow \sin x<\frac{1}{2}\Rightarrow x\in (0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$

Therefore, Answer is $x\in (0,\frac{\pi }{6})\cup (\frac{5\pi }{6},2\pi )$