The general solution of $sin 4 x cos 2 x = cos 5 x sin x$ is

{\frac{n π}{3}} \cup {(2 n + 1) \frac{π}{2}}, n \in Z

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{\frac{n π}{2}} \cup {(4 n + 1) \frac{π}{6}}, n \in Z

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{\frac{n π}{3}} \cup {(2 n + 1) \frac{π}{6}}, n \in Z

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{\frac{n π}{2}} \cup {(4 n + 1) \frac{π}{6}}, n \in Z

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Solution

The correct option is A ${\frac{n π}{3}} \cup {(2 n + 1) \frac{π}{2}}, n \in Z$
$sin 4 x cos 2 x = cos 5 x sin x \Rightarrow 2 sin 4 x cos 2 x = 2 cos 5 x sin x$
$\Rightarrow sin 6 x + sin 2 x = sin 6 x - sin 4 x$
$\Rightarrow sin 2 x + sin 4 x = 0$
$\Rightarrow 2 sin 3 x cos x = 0$
$\Rightarrow sin 3 x = 0 or cos x = 0$
$i . e ., 3 x = n π or x = (2 n + 1) \frac{π}{2}$
$\Rightarrow x = \frac{n π}{3} or x = (2 n + 1) \frac{π}{2}, n \in Z$

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