Let $2 {sin}^{2} x + 3 sin x - 2 > 0$ and $x^{2} - x - 2 < 0$ ( $x$ is measured in radians). Thcn $x$ lies in the interval

(\frac{π}{6} . \frac{5 π}{6})

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(- 1, \frac{5_{π}}{6})

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(- 1, 2)

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(\frac{π}{6}, 2)

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Solution

The correct option is D $(\frac{π}{6}, 2)$
first inequality gives
$(2 sin x - 1) (sin x + 2) > 0$
$(sin x - \frac{1}{2}) > 0$
$\frac{Π}{6} < x < \frac{5 Π}{6}$
second inequality gives
$(x + 1) (x - 2) < 0$
$- 1 < x < 2$
common part is $\frac{Π}{6} < x < 2$

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