Let 2{sin ^2}x + 3sin x - 2 0 and {x^2} | vedclass

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Question

(d) $2\,{\sin ^2}x + 3\,\sin x - 2 > 0$

$2\,{\sin ^2}x + 4\,\sin x - \sin x - 2 > 0$

$2\,\sin x\,(\sin x + 2) - 1\,(\sin x + 2) > 0$

Vedclass · Accepted Answer

$\left( {\frac{\pi }{6},\;2} \right)$

Vedclass · Answer

(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 $(ii)$, $x \in (\pi /6,\,\,2)$.

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

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