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

$\sin x - 3\sin2x + \sin 3x = \cos x -3\cos2x + \cos 3x$ 
OR 
$\sin x + \sin 3x - 3\sin 2x = \cos x + \cos 3x - 3\cos 2x$ 
$\implies 2\sin 2x \cdot \cos x - 3\sin 2x = 2\cos 2x \cdot \cos x - 3\cos 2x $ 
$\implies \{\sin 2x - \cos 2x\} \{ 2\cos x - 3\} = 0 $ 
$\implies \{\sin 2x - \cos 2x\} = 0 $ since $\cos x \neq 3/2 $ 
$\implies \cos 2x = \cos (\pi/2 - 2x) $ 
$\implies 4x = 2n\pi + \pi/2 $
$\implies x = \pi/8 \implies n\pi/2 \pm {\pi\over 8} $ in general form . 
