Trigonometric equation, missing some solutions I'm missing part of the answer, and I'm not quite sure why. The given answer doesn't even seem to hold... 
Solve for x: $$\tan 2x = 3 \tan x $$
First some simplifications:
$$\tan 2x = 3 \tan x $$
$$\tan 2x - 3 \tan x  = 0$$
$$\frac{\sin 2x}{\cos 2x} - \frac{3 \sin x}{\cos x} = 0$$
$$\frac{2 \sin x \cos^2x - 3 \sin x \cos 2x}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(2 \cos^2x - 3 \cos 2x)}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(2 \cos^2x - 3 (\cos^2 x - \sin^2 x))}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(2 \cos^2x - 3\cos^2 x + \sin^2 x)}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(\sin^2 x - \cos^2 x)}{\cos 2x \cos x} = 0$$
$$\frac{\sin x(\sin^2 x - \cos^2 x)}{(\sin^2 x - \cos^2 x) \cos x} = 0$$
$$\frac{\sin x}{\cos x} = 0$$
Looks much simpler. Now solving for x, since $\frac{\sin x}{\cos x} = 0 $ when $\sin x = 0$ and $\sin x = 0$ for every half rotation, the answer must be $k\pi$.
Alas, according to my answer sheet, I'm missing two values: $\frac{\pi}{6} + k\pi$ and $\frac{5\pi}{6} + k\pi$. But since $\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})} = \frac{\sqrt3}{3}$, I'm not sure where these answers come from. 
Furthermore, this is the kind of mistake I'm making all over these exercises, I'd like to avoid that, but how can I be sure I have ALL the answers needed? 

Setting $\tan x=t$
we have $$\frac{2t}{1-t^2}=3t\iff2t=3t(1-t^2)\iff t(2-3+3t^2)=0$$
If $t=0,\tan x=0, x=n\pi$ where $n$ is any integer
$2-3+3t^2=0\iff 3t^2=1\implies\cos2x=\dfrac{1-t^2}{1+t^2}=\dfrac12=\cos\dfrac\pi3$
$\implies2x=2m\pi\pm\dfrac\pi3$ where $m$ is any integer
