The general solution of the equation sin⁡x−3sin⁡2x+sin⁡3x=cos⁡x−3cos⁡2x+cos⁡3x is

A
$n π + \frac{π}{4}$
B
$\frac{n π}{2} + \frac{π}{4}$
C
$n π + \frac{π}{8}$
D
$\frac{n π}{2} + \frac{π}{8}$

Solution:

$\sin x + \sin 3 x - 3 \sin 2 x$ $= \cos x + \cos 3 x - 3 \cos 2 x$

$\begin{array}{l} \Rightarrow 2 \sin 2 x \cos x - 3 \sin 2 x = 2 \cos 2 x \cos x - 3 \cos 2 x \\ \Rightarrow (2 \cos x - 3) (\sin 2 x - \cos 2 x) = 0 \\ \Rightarrow 2 \cos x \neq 3 ∴ \tan 2 x = 1 \\ \Rightarrow \tan 2 x = \tan \frac{π}{4} ∴ 2 x = n π + \frac{π}{4} \Rightarrow x = \frac{n π}{2} + \frac{π}{8} \end{array}$

The general solution of the equation $\sin x - 3 \sin 2 x + \sin 3 x = \cos x - 3 \cos 2 x + \cos 3 x$ is

Solution:

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