Gfilui

I'm given this trigonometric equation,
$$tan x+tan 2x=1-tan xtan 2x$$
I rewrote it as $$dfractan x+tan 2x1-tan xcdottan 2x=1$$
Using the identity, $$tan (A+B)=dfractan A+tan B1-tan Acdottan B$$ I simplified my equation as $$tan (x+2x)=1$$ which implies $$3x=dfracpi4+npi$$
So, $$x=dfracpi12+ncdotdfracpi3$$
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?

Note that the statement $$dfractan x+tan 2x1-tan xcdottan 2x=1$$ is valid only if the denominator $$1-tan xcdottan 2x ne 0$$

We need to exclude the values for which $$ tan xcdottan 2x=1$$ from the solutions that you have found.

This latest equation is not hard to solve and I let you handle it.

i would use that $$tan(2x)=frac2tan(x)1-tan^2(x)$$
Then you will get

$$tan(x)+frac2tan(x)1-tan^2(x)=1-frac2tan^2(x)1-tan^2(x)$$

now substitute $tan(x)=t$

You can divide with $tan x tan 2x-1$ iff $tan x tan 2xne 1$

so $$2tan^2xover 1-tan^2x neq 1implies tan x neq pmsqrt3over 3 $$

so $xneq pm pi over 6+pi cdot k;; k in mathbbZ$.

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

$$begincases
x = infty, & text$forall n$ a multiple of $3n-1$ \
x = dfrac pi (4n+1)12, & text$forall n$ not a multiple of $3n-1$
endcases$$

The reason why Wolfram Alpha wrote its general solutions as $$x = n pi - dfrac 7 pi12 text and x = n pi - dfrac 11 pi12$$ 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.