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Let $2{\sin ^2}x + 3\sin x - 2 > 0$ and ${x^2} - x - 2 < 0$ ($x$ is measured in radians). Then $x$ lies in the interval
$\left( {\frac{\pi }{6},\;\frac{{5\pi }}{6}} \right)$
$\left( { - 1,\;\frac{{5\pi }}{6}} \right)$
$( - 1,\;2)$
$\left( {\frac{\pi }{6},\;2} \right)$
Solution
(d) $2\,{\sin ^2}x + 3\,\sin x – 2 > 0$
$2\,{\sin ^2}x + 4\,\sin x – \sin x – 2 > 0$
$2\,\sin x\,(\sin x + 2) – 1\,(\sin x + 2) > 0$
$\,(\sin x + 2)\,(2\sin x – 1) > 0$
$2\,\sin x – 1 > 0\,\, \Rightarrow \,\,\sin x > 1/2$
$x > \pi /6\,\, \Rightarrow \,\,x \in \,\,(\pi /6,\,\,\infty )…..(i)$
and ${x^2} – x – 2 < 0\, \Rightarrow \,{x^2} – 2x + x – 2 < 0$
$x\,(x – 2) + 1\,(x – 2) < 0$
$(x + 1)\,(x – 2) < 0\, \Rightarrow \,\,x \in ( – 1,\,\,2)…..(ii)$
From $(i)$ and $(ii)$, $x \in (\pi /6,\,\,2)$.
