Where trigonometry root $ 2\pi n $ comes from I've and expression like:
$$ \tan^2(x) = \frac{1-\cos^3(x)}{1-\sin^3(x)} $$
and i found roots:
$$ x = \pi n + \frac{\pi}{4} $$
$$ x = (-1)^n * arcsin \frac{\sqrt 2 - 1}{\sqrt 2} + \pi n - \frac{\pi}{4}  $$
but there should be one more which equals to $ 2 \pi n $. Where it comes from as i'm having the following expression:
$$ (\sin(x) - \cos(x)) (\sin(x) + \cos(x) + \sin(x)\cos(x)) = 0 $$
From the first multiplier i have the first root, and from the second i have two, but one is below -1 which is not satisfying requirements.
UPD: I only interested in the root of $ 2\pi n $. I want to see how could i get it, not just guessing that it suits.
 A: Hint:
Writing $\sin x=s,\cos x=c$
$$\dfrac{(1-c)(1+c+c^2)}{(1-s)(1+s+s^2)}=\dfrac{(1-c)(1+c)}{(1-s)(1+s)}$$
If $1-s=0\iff s=1,$ both sides are undefined $\implies s\ne1$
If $1-c=0\iff\cos x=1,x=2m\pi$ where $m$ is any integer
Else
$$\dfrac{1+c+c^2}{1+s+s^2}=\dfrac{1+c}{1+s}$$
$$\iff(s-c)(s+c+sc)=0$$
Method$\#1:$
Use Weierstrass Substitution 
Method$\#2:$ 
Squaring & on rearrangement  we get $$\sin^22x-4\sin2x-1=0$$
$$\sin2x=2\pm\sqrt5$$
As for real $x,\sin2x\le1,$  $$\sin2x\ne2+\sqrt5$$
Now discard the extraneous root.
A: Consider $$
\tan^2(x)=0\Leftrightarrow \tan(x)=0\Leftrightarrow x=\pi n \text{ for }n\in\mathbb Z.
$$
So you see that your formula is wrong since
$$
\tan^2(\pi)=0\neq 2=\frac{1-(-1)^3}{1-0^3}=\frac{1-\cos^3(\pi)}{1-\sin^3(\pi)}.
$$
The right formula, using $\tan=\frac{\cos}{\sin}$ and $\cos^2+\sin^2=1$, is given by
$$
\tan^2(x)=\frac{\sin^2(x)}{\cos^2(x)}=\frac{1-\cos^2(x)}{1-\sin^2(x)}
$$
Now you can also see
$$
\tan^2(x)=0\Leftrightarrow 1-\cos^2(x)=0\Leftrightarrow \cos(x)\in\{-1,1\}\Leftrightarrow x=\pi n\text{ for }n\in\mathbb Z.
$$
But if you are interested in the roots of $f(x):=\frac{1-\cos^3(x)}{1-\sin^3(x)}$ you get
$$
f(x)=0\Leftrightarrow 1-\cos^3(x)=0\Leftrightarrow \cos^3(x)=1\Leftrightarrow \cos(x)=1\Leftrightarrow x=2\pi n\text{ for }n\in\mathbb Z.
$$
Since $1-\sin^3(2\pi n)=1\neq 0$ for $n\in\mathbb Z$ we deduce that $2\pi n$ for $n\in\mathbb Z$ are all roots of $f$.
