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?
 A: When you arrive at
$$
-\tan^2x \frac{2\tan x}{1-\tan^2x} = \frac{2\tan x}{1-\tan^2x}
$$
you can't multiply both sides by $\frac{1-\tan^2x}{2\tan x}$, because this operation is allowed only when the multiplier is nonzero.
You should rather move everything to the right-hand side and collect terms, getting
$$
0=\frac{2\tan x}{1-\tan^2x}(\tan^2x+1)
$$
Since $\tan^2x+1\ne0$, you get
$$
\tan x=0
$$
and so $x=k\pi$.

You can also observe that $\tan3x=\tan x$ means
$$
3x=x+k\pi
$$
so
$$
x=k\frac{\pi}{2}
$$
provided $\tan x$ and $\tan3x$ exist, which is not the case when $k$ is odd. Thus the solutions are $x=(2h)\frac{\pi}{2}=h\pi$ ($h$ integer).
A: You can't obviously cancel in the last (or one before the last, in fact) step, thus you must have
$$\frac{2\tan x}{1-\tan^2x}=0\iff \tan x=0\iff x=k\pi\;,\;\;k\in\Bbb Z.$$
A: 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 $x = k\pi$".
=====
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 $x = \frac n \pi$.
A: Why do you try to make things more complex than they are? Formally,
\begin{align*}\tan x=\tan 3x&\iff 3x\equiv x\mod \pi \\
&\iff 2x\equiv0\mod \pi 
\iff x\equiv 0\mod\frac\pi2.
\end{align*}
However, $\tan x$ is defined if and only if $x\not\equiv\dfrac\pi2\mod\pi$, so the effective solutions are
$$x\equiv 0\mod \pi.$$
A: When you have this:
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$
It it incorrect to have this next:
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$
Just like when you have (for example) $3 \cdot 2x = x$, it is incorrect to then divide both sides by $x$ and end up with $6 = 1$ (you lose the solution $x = 0$ when you do this).
The correct next step is to subtract $\dfrac{2\tan x}{1-\tan^2 x}$ from both sides and continue as follows:
\begin{align}
  -\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} &= \frac{2\tan{x}}{1-\tan^2{x}}\\[0.3cm]
  -\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} - \frac{2\tan{x}}{1-\tan^2{x}}  &= 0\\[0.3cm]
  \frac{2\tan{x}}{1-\tan^2{x}} \left(-\tan^2 x - 1\right) &= 0
\end{align}
So then we have the following two equations:
$$\frac{2\tan x}{1-\tan^2 x} = 0 \qquad \text{or} \qquad -\tan^2 x - 1 = 0$$
The first equation is trivial.  The second one is perhaps even more so because the second one has no solutions.
A: Why do you have an error?
If you divide though (by $\frac {2\tan x}{1-\tan^2x} = \tan 2x$ in this case), you must check that whatever you are dividing by does not equal 0.  Or, make a note to review the case that it does.
$\tan^2 x = -1$ or $\frac {2\tan x}{1-\tan^2x} = 0$
An alternatives you could have taken:
$0 = (1 - \tan^2 x)\frac {2\tan x}{1-\tan^2x}\\
0 = \sec^2 x \tan 2x$
But you are working a little bit too hard.
$\tan x = \frac {\tan x + \tan 2x}{1-\tan x\tan 2x}$
Rather than expand $\tan 2x$ right now, lets simplify a bit first
$\tan x- \tan^2 x \tan 2x = \tan x + \tan 2x\\
0 = \tan 2x (1+\tan^2 x)\\
0 = \tan 2x(\sec^2 x)$
Now, let's be even a little bit stupider.
$\tan x = \tan (x + k\pi)\\
\tan (x + k\pi) = \tan 3x\\
\tan^{-1}(\tan (x + k\pi)) = \tan^{-1}(\tan 3x)\\
x + k\pi = 3x\\
2x = k\pi$
