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

\begin{align}
\sin x - 3\sin 2x + \sin 3x &= \cos x - 3\cos 2x + \cos3x\\
2\sin 2x \cos x  - 3\sin 2x &= 2\cos 2x  \cos x - 3\cos 2x \\
\sin 2x (2 \cos x  - 3) &= \cos 2x  (2\cos x - 3) \\
\end{align}
Now if $2 \cos x  - 3=0$ then $\cos x=\frac 32$ which is not possible, hence as $2 \cos x  - 3 \neq0$ then we divide both sides by $2 \cos x  - 3$ to obtain $\sin 2x = \cos 2x$ which is solvable.
NOTE I used sum-to-product relations from here.
