If $2 {cos}^{2} x < 2 - 3 cos x$ and $, x^{2} < 4 x + 12,$ then x belongs to

(- 2, 6)

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(- 2, \frac{1}{2})

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(\frac{π}{3}, \frac{π}{2}) \cup (\frac{3 π}{2}, 2 π)

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(- 2, - \frac{π}{3}) \cup (\frac{π}{3}, \frac{5 π}{3})

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Solution

The correct option is D $(- 2, - \frac{π}{3}) \cup (\frac{π}{3}, \frac{5 π}{3})$
$2 {cos}^{2} x < 2 - 3 c o x \Rightarrow 2 {cos}^{2} x + 3 cos x - 2 < 0 (2 cos x - 1) (cos x + 2) < 0 cos x ϵ (- 2, \frac{1}{2}) \Rightarrow cos x ϵ (- 1, \frac{1}{2}) A s - 1 \leq cos x \leq 1 ∴ x ϵ (2 m π + (- \frac{π}{3}), 2 m π + \frac{π}{3}) - - (1)$
And $x^{2} < 4 x + 12 \Rightarrow x^{2} - 4 x - 12 < 0 \Rightarrow (x + 2) (x - 6) < 0 \Rightarrow x ϵ (- 2, 6) - - (2)$
Taking intersection of (1) and (2), we get
$x ϵ (- 2, - \frac{π}{3}) \cup (\frac{π}{3}, \frac{5 π}{3})$

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