Since $1 + \tan x = 2 + \sqrt{2}$ and $1 - \tan x = -\sqrt{2}$, then
$$\tan\left(\frac{\pi}{4}-x\right) = \frac{1 - \tan x}{1 + \tan x} = \frac{-\sqrt{2}}{2 + \sqrt{2}} = -\frac{1}{\sqrt{2}+1}= -\frac{1}{\tan x}.$$
Therefore,
$$1 + \tan x\, \tan\left(\frac{\pi}{4}-x\right) = 0,$$
which implies
$$\frac{1 + \tan x\tan\left(\frac{\pi}{4}-x\right)}{\tan x - \tan\left(\frac{\pi}{4}-x\right)} = 0,$$
or,
$$\cot\left(x + \left(\frac{\pi}{4}-x\right)\right) = 0,$$
that is,
$$\cot\left(2x - \frac{\pi}{4}\right) = 0.$$
One of the solutions to this equation is such that $2x - \frac{\pi}{4} = \frac{\pi}{2}$, or $x = \frac{3\pi}{8}$. So all solutions are of the form $\frac{3\pi}{8} + n\pi$, where $n$ ranges over the integers. Since we seek those solutions in $(0, \frac{\pi}{2})$, the only root of the original equation is $x = \frac{3\pi}{8}$.