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

(π / 6, 5 π / 6)

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(- 1, 5 π / 6)

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(- 1. 2)

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(π / 6, 2)

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Solution

The correct option is D $(π / 6, 2)$
$x^{2} - x - 2 < 0$
$\Rightarrow (x + 1) (x - 2) < 0$
i.e $x ϵ (- 1, 2) ⟶ (1)$
Also given,
$2 s i n^{2} x + 3 s i n x - 2 > 0$
$\Rightarrow (2 s i n x - 1) (s i n x + 2) > 0$
we know $- 1 \leq s i n x \leq 1 \Rightarrow s i n x + 2 > 0$
$\Rightarrow 2 s i n x - 1 > 0$
$\Rightarrow \frac{1}{2} < s i n x \leq 1 ⟶ (2)$
considering (1) and (2), solution is,
$x ϵ (\frac{π}{6}, 2)$
Hence, option 'D' is correct.

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