I know that $\cos 3x = 4\cos^3 x - 3\cos x,$ so I can substitute that into the given equation and simplify. I get $\frac{4\cos^3 x - 3\cos x}{\cos x} = \frac{1}{3},$ or $4\cos^2 x - 3 = \frac{1}{3}.$ Solving for $\cos^2 x,$ I get $\cos^2 x = \frac{10}{12} = \frac{5}{6}.$ Now, I also know that $\sin 3x = 3\sin x - 4\sin^3 x,$ so I can plug that into the expression I want to find. I get $\frac{\sin 3x}{\sin x} = \frac{3\sin x - 4\sin^3 x}{\sin x} = 3 - 4\sin^2 x.$ To find $\sin^2 x,$ I can use the identity $\sin^2 x + \cos^2 x = 1.$
[* no more bullets] Since $\cos^2 x = \frac{5}{6},$ I get $\sin^2 x = 1 - \frac{5}{6} = \frac{1}{6}.$
[* no more bullets] Therefore, $\frac{\sin 3x}{\sin x} = 3 - 4\cdot \frac{1}{6} = 3 - \frac{2}{3} = \frac{7}{3}.$


If $\frac{\cos 3x}{\cos x} = \frac{1}{3},$ then determine $\frac{\sin 3x}{\sin x}.$