Solve the equation: $ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$ How do I solve this equation:
$$ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$$

First, we may use some identities that you can find here. That's how we proceed
$$\begin{align}
\sin x - 3\sin 2x + \sin 3x &= (\sin x + \sin 3x) - 3\sin 2x \\
&=2\sin2x\cos x-3\sin2x \\
&=\sin2x(2\cos x - 3)
\end{align}$$
$$\begin{align}
\cos x - 3\cos 2x + \cos 3x &= (\cos x + \cos 3x) - 3\cos 2x \\
&=2\cos2x\cos x-3\cos2x \\
&= \cos2x(2\cos x - 3)
\end{align}$$
and then we can get
$$
\sin2x(2\cos x - 3) = \cos2x(2\cos x - 3) \\
(\sin2x-\cos2x)(2\cos x - 3) =0
$$
and hence
$$\begin{cases}
\sin2x-\cos2x=0 & \to \sqrt{2}\sin(2x-\frac{\pi}{4})=0\\
\text{or} \\
\cos x=\frac{3}{2} & \to \text{impossible}
\end{cases}$$
Where in the first implication, I used the identity mentioned here. This finally leads to
$$2x - \frac{\pi}{4} = n\pi, \qquad n=0,\pm1,\pm2,...$$
Or equivalently
$$x=\frac{\pi}{8} + n\frac{\pi}{2}, \qquad n=0,\pm1,\pm2,...$$
