General values of $x$ for which $sin 2 x + cos x = 0$ is/are :

\frac{2 n π}{3} + \frac{π}{6}

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\frac{2 n π}{3} - \frac{π}{6}

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2 n π + \frac{π}{2}

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2 n π - \frac{π}{2}

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Solution

The correct option is D $2 n π - \frac{π}{2}$
$sin 2 x + cos x = 0 \Rightarrow cos x = - sin 2 x \Rightarrow cos x = cos (\frac{π}{2} + 2 x) \Rightarrow x = 2 n π \pm (\frac{π}{2} + 2 x), n \in Z$

When we consider the positive sign, we have :
$\Rightarrow x = 2 n π + (\frac{π}{2} + 2 x) \Rightarrow - x = 2 n π + \frac{π}{2} \Rightarrow x = 2 m π - \frac{π}{2} (m = - n)$

When we consider the negative sign, we have :
$\Rightarrow x = 2 n π - (\frac{π}{2} + 2 x) \Rightarrow x = \frac{2 n π}{3} - \frac{π}{6}$

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