Question

Set $2si{n}^{2}x+3sinx-2>0$ and $x^2-x-2<0$ (x is measured in radians). Then x lies in the interval

• A. $(\pi /6,5\pi /6)$
• B. $(-1,5\pi /6)$
• C. $(-1.2)$
• D. $(\pi /6,2)$

Answer 1

The correct option is D $(\pi /6,2)$

$x^2-x-2<0$
$\Rightarrow (x+1)(x-2)<0$
i.e $xϵ(-1,2)⟶\left(1\right)$
Also given,
$2si{n}^{2}x+3sinx-2>0$
$\Rightarrow (2sinx-1)(sinx+2)>0$
we know $-1\le sinx\le 1\Rightarrow sinx+2>0$
$\Rightarrow 2sinx-1>0$
$\Rightarrow \frac{1}{2}<sinx\le 1⟶\left(2\right)$
considering (1) and (2), solution is,
$xϵ(\frac{\pi }{6},2)$
Hence, option 'D' is correct.