solve $\tan{x} = \tan{3x}$ I'm asked to solve $\tan{x} = \tan{3x}$
Here's my attempt: 
$$\tan{x} = \tan{3x}$$
$$\tan{x} = \tan{(x + 2x)}$$
$$\tan{x} = \frac{\tan{x} + \tan{2x}}{1-\tan{x}\tan{2x}}$$
Recall the identity:
$$\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}$$
So then we have:
$$\tan{x} = \frac{\tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}}{1-\tan{x}\frac{2\tan{x}}{1-\tan^2{x}}}$$
$$\tan{x} - \tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$
$$\tan^2{x} = -1$$
This does obviously not compute. Why is my way wrong and how can I go about solving it?

Your proof is fine up to 
$-\tan^2{(x)}(\frac{2\tan{x}}{1-\tan^2{x}}) = \frac{2\tan{x}}{1-\tan^2{x}}$
Then you do $-\tan^2{x} \cdot (\frac{2\tan{x}}{1-\tan^2{x}}) \cdot (\frac{1-\tan^2{x}}{2\tan{x}}) = 1$.
But you can onlty do that if $\frac{2\tan{x}}{1-\tan^2{x}} \ne 0$.
So you need to say:
"Assume $\frac{2\tan{x}}{1-\tan^2{x}}\ne 0$ then
"$-\tan^2{x} \cdot (\frac{2\tan{x}}{1-\tan^2{x}}) \cdot (\frac{1-\tan^2{x}}{2\tan{x}}) = 1$
"$-\tan^2{x} = 1$.
"But this is impossible.  
"So $\frac{2\tan{x}}{1-\tan^2{x}} = 0$"
And go on from there:
"So $\tan x = \sin x/\cos x = 0$.
"So $x = k\pi$".
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Or you can note $tan z = tan x \iff z = x + k\pi$.
So $3x = x + k\pi$ so $2x = k\pi$ so $x = \frac k 2 \pi$.  But $\tan \frac k 2 \pi; k$ odd is undefined so $x = \frac k 2 \pi; k$ even or in other words $x = \frac n \pi$.
