<s>[INST] Solving a trig. equation Solve for x $$\tan 2x = 3 \tan x$$
Well, I'm stuck with this one. This is how far I got. 
$$\tan 2x = 3 \tan x$$
$$\frac{2 \tan x}{1 - \tan^2 x} = 3 \tan x$$
$$2 \tan x = 3 \tan x - 3 \tan^3 x$$
$$3 \tan^3 x = \tan x$$
$$3 \tan^3 x - \tan x = 0$$
$$\tan x (3 \tan^2 x - 1) = 0$$
So, then a part of the solution is $\tan x = 0$, so $x = k\pi$
This leaves $\tan^2x = 1$. I thought to take the root of both sides, leaving me with $\tan x = 1$. So I figured $x = \frac{1}{4}\pi + k\pi$ so we the first and third quadrants, where the tangent is positive.
But, to my deep deep sadness, this last part turned out to be wrong. 
What am I missing here?
 [/INST] The equation $3\tan^3 x - \tan x = 0$ factors as $\tan x(\color{red}{3}\tan^2 x - 1) = 0$ (you forgot the $3$). 
Hence, $\tan x = 0$ or $\tan x = \pm \dfrac{1}{\sqrt{3}}$. Both of these should be easy to solve.
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