If $\cos x + \cos 2x + \cos 3x = 3,$ then find $\sin x + \sin 2x + \sin 3x.$
Remember that $\cos \theta \le 1$ for all angles $\theta.$  So, the only way that the equation
\[\cos x + \cos 2x + \cos 3x = 3\]can holds is if $\cos x = \cos 2x = \cos 3x = 1.$  In such a case,
\[\sin^2 x = 1 - \cos^2 x = 0,\]so $\sin x = 0,$ which means $x$ is a multiple of $\pi.$  Therefore, $\sin x + \sin 2x + \sin 3x = \boxed{0}.$  (Note that we can take $x = 0.$)