Let $2 s i n^{2} x + 3 s i n x - 2$ >0 and $x^{2} - x - 2$ < 0 (x is measured in radians). Then x lies in the Interval

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Solution

The correct option is D
$2 s i n^{2} x + 3 s i n x - 2$ >0 $\Rightarrow (2 s i n x - 1) (s i n x + 2)$ >0 $\Rightarrow 2 s i n x - 1$ >0
$\Rightarrow s i n x$ > $\frac{1}{2} \Rightarrow x ϵ (\frac{π}{6}, 5 \frac{π}{6})$
Also $x^{2} - x - 2$ <0 $\Rightarrow (x - 2) (x + 1)$ <0 $\Rightarrow$ -1 < x < 2.
$∴ x m u s t l i e i n (\frac{π}{6}, 2)$

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