What is the first step to solving $\cos3x - \sin x = \sqrt{3}(\cos x - \sin 3x)$? My calculus BC teacher has given us some trig "review". 
$$\cos3x - \sin x = \sqrt{3}(\cos x - \sin 3x).$$
How do I get rewrite the cos3x and sin3x? Do I just use sum and difference, because it makes everything really complicated. 
can someone teach me how to format this correctly... sorry I am a new user

$\cos3x - \sin x = \sqrt{3}(\cos x - \sin 3x)$  
$\implies \cos3x+\sqrt3\sin3x=\sqrt3\cos x+\sin x$  
Now divide both sides by 2 to get :  
$\frac{1}{2}\cos3x+\frac{\sqrt3}{2}\sin3x=\frac{\sqrt3}{2}\cos x+\frac{1}{2}\sin x$  
$\implies \cos \frac{\pi}{3} \cos 3x + \sin \frac{\pi}{3} \sin 3x=\cos \frac{\pi}{6} \cos x + \sin \frac{\pi}{6} \sin x$  
$\implies \cos(3x-\frac{\pi}{3})=\cos(x-\frac{\pi}{6})$  
$\implies 3x-\frac{\pi}{3}=\pi k \pm (x-\frac{\pi}{6}),  k\in\mathbb{Z}$  
$\implies x=\frac{\pi}{2} \left( k+\frac{1}{6}\right)  OR $  $ $ $ x=\frac{\pi}{4} \left( k+\frac{1}{2}\right)$
