General solutions of the equation $cos 3 x = sin 2 x$ is

{\frac{2 n π}{5} - \frac{π}{10}} \cup {2 n π + \frac{π}{2}}, n \in Z

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{\frac{2 n π}{5} - \frac{π}{10}} \cup {2 n π - \frac{π}{2}}, n \in Z

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{\frac{2 n π}{5} + \frac{π}{10}} \cup {2 n π + \frac{π}{2}}, n \in Z

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{\frac{2 n π}{5} + \frac{π}{10}} \cup {2 n π - \frac{π}{2}}, n \in Z

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Solution

The correct option is D ${\frac{2 n π}{5} + \frac{π}{10}} \cup {2 n π - \frac{π}{2}}, n \in Z$
$cos 3 x = sin 2 x \Rightarrow cos 3 x = cos (\frac{π}{2} - 2 x) \Rightarrow 3 x = 2 n π \pm (\frac{π}{2} - 2 x) [∵ cos θ = cos α \Rightarrow θ = 2 n π \pm α] \Rightarrow 3 x = 2 n π + (\frac{π}{2} - 2 x) \Rightarrow x = \frac{2 n π}{5} + \frac{π}{10}, where n \in Z$
or
$3 x = 2 n π - (\frac{π}{2} - 2 x) \Rightarrow x = 2 n π - \frac{π}{2}, where n \in Z ∴ x = {\frac{2 n π}{5} + \frac{π}{10}} \cup {2 n π - \frac{π}{2}}, n \in Z$

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[0, 2 π]

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