Solution of trigonometric equation avoiding the undefined case I'm given this trigonometric equation,
$$\tan x+\tan 2x=1-\tan x\tan 2x$$
I rewrote it as $$\dfrac{\tan x+\tan 2x}{1-\tan x\cdot\tan 2x}=1$$
Using the identity, $$\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\cdot\tan B}$$ I simplified my equation as $$\tan (x+2x)=1$$ which implies $$3x=\dfrac{\pi}{4}+n\pi$$
So, $$x=\dfrac{\pi}{12}+n\cdot\dfrac{\pi}{3}$$
Where, $n$ is an integer.
However, when $n=5$, $\tan (2x)$ is undefined. 
This is causing a problem.
WolframAlpha gave solutions which avoids my case.
How do I come to the solution which WolframAlpha gives?
 A: Note that the statement $$\dfrac{\tan x+\tan 2x}{1-\tan x\cdot\tan 2x}=1$$ is valid only if the denominator $$1-\tan x\cdot\tan 2x \ne 0$$
We need to exclude the values for which $$ \tan x\cdot\tan 2x=1$$ from the solutions that you have found.
This latest equation is not hard to solve and I let you handle it.
A: i would use that $$\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$$
Then you will get
$$\tan(x)+\frac{2\tan(x)}{1-\tan^2(x)}=1-\frac{2\tan^2(x)}{1-\tan^2(x)}$$
now substitute $\tan(x)=t$
A: You can divide with $\tan x \tan 2x-1$ iff $\tan x \tan 2x\ne 1$
so $${2\tan^2x\over 1-\tan^2x} \neq 1\implies \tan x \neq \pm{\sqrt{3}\over 3} $$
so $x\neq \pm {\pi \over 6}+\pi \cdot k;\; k \in \mathbb{Z}$.
A: The particular solutions ($0<x<2 \pi$) are $\dfrac {\pi} {12}$ and $\dfrac {5 \pi} {12}$, but when you work out both Wolfram Alpha's and the OP's solution, both general solutions come out to $$\dfrac {\pi (4n+1)}{12}$$
The problem with that is you have to have a restriction for $n=2, n=5, n=8$ and $n=11$, which causes $\tan 2x$ to be undefined.  To restrict those values, we can write 
$$\begin{cases}
x = \infty, & \text{$\forall n$ a multiple of $3n-1$} \\
x = \dfrac {\pi (4n+1)}{12}, & \text{$\forall n$ not a multiple of $3n-1$}
\end{cases}$$
The reason why Wolfram Alpha wrote its general solutions as $$x = n \pi - \dfrac {7 \pi}{12} \text { and } x = n \pi - \dfrac {11 \pi}{12}$$ are because the negative angles are complementary to the positive ones, i.e. $\frac {\pi} {12}$ and $\frac {5 \pi} {12}$ counterclockwise around the unit circle are equivalent to $-\frac {7 \pi} {12}$ and $-\frac {11 \pi} {12}$ clockwise around the unit circle.  They also avoid the the undefined solutions altogether.
