Question

If $2{\sin }^{2}x+3\sin x-2>0$ and $x^2-x-2<0$, then $x$ lies in the interval

• A. $(\frac{\pi }{6},\frac{5\pi }{6})$
• B. $(-1,\frac{5\pi }{6})$
• C. $(-1,2)$
• D. $(\frac{\pi }{6},2)$

Answer 1

The correct option is D $(\frac{\pi }{6},2)$

$x^2-x-2<0\Rightarrow (x-2)(x+1)<0\Rightarrow -1<x<2\cdots \left(i\right)$
Also,
$2{\sin }^{2}x+3\sin x-2>0$
$\Rightarrow (2\sin x-1)(\sin x+2)>0$
$\Rightarrow 2\sin x-1>0[\because -1\le \sin x\le 1]$
$\Rightarrow \sin x>\frac{1}{2}$
Now using the graph

$\Rightarrow x\in (\frac{\pi }{6},\frac{5\pi }{6})\cdots \left(ii\right)$

From equation $\left(i\right)$ and $\left(ii\right)$, we get
$x\in (\frac{\pi }{6},2)$