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Left Hand Limit

Let 2 sin 2 x...

Question

Let $2 s i n^{2} x + 3 s i n x - 2 > 0$ and $x^{2} - x - 2 < 0$ . Then x lies in the interval

(\frac{π}{6}, \frac{5 π}{6})

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(- 1, 2)

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(\frac{π}{6}, 2)

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(- 1, \frac{5 π}{6})

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Solution

The correct option is C $(\frac{π}{6}, 2)$

$(2 s i n x - 1) (s i n x + 2) > 0 \Rightarrow s i n x > \frac{1}{2} \Rightarrow x \in (\frac{π}{6}, \frac{5 π}{6}) x^{2} - x - 2 > 0 \Rightarrow (x - 2) (x + 1) < 0 \Rightarrow x \in (- 1, 2)$
$∴$ common value of $x \in (\frac{π}{6}, 2)$

Suggest Corrections

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2 {sin}^{2} x + 3 sin x - 2 > 0

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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

2 {sin}^{2} x + 3 sin x - 2 > 0

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