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Sinx = 2sinx cos2x is satisfied if x belongs to

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$\{nπ \pm \frac{π}{6}\} \cup \{nπ\}, n \in z$

$\{nπ \pm \frac{π}{3}\} \cup \{\frac{nπ}{3}\}, n \in z$

$\{nπ \pm \frac{π}{2}\} \cup \{\frac{nπ}{2}\}, n \in z$

$\{nπ \pm \frac{π}{4}\} \cup \{\frac{nπ}{6}\}, n \in z$

answer is D.

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Detailed Solution

$\sin x = 2 \sin {xcos}^{2} x 2 \sin x \cos 2 x - \sin x = 0 \sin x (2 \cos 2 x - 1) = 0 \sin x = 0, \cos 2 x = \frac{1}{2} x = nπ 2 x = 2 nπ \pm \frac{π}{3} x = nπ, nπ \pm \frac{π}{6}$

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