Oh, I see. Maybe now is clear: $2\alpha$ denotes just a particular solution in $[0,\pi]$ for the equation $\tan(2x)=-\frac{1}{4k}$, in general I must find a solution in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ adding the period $h\pi$ ($h\in\mathbb{Z}$) because tan is defined there, but here $2x\in[0,\pi]$ and so I must restrict the interval where I'm searching the solutions to $[0,\pi] \setminus \{\frac{\pi}{2}\}$. So the only possibilities is that $2\alpha \in \left[0,\frac{\pi}{2}\right)$ or $2\alpha \in \left(\frac{\pi}{2},\pi\right]$; for what you've said in the answers, the only possibility is $2\alpha \in \left(\frac{\pi}{2},\pi\right]$ because is the only one that is coherent with the sign of $-\frac{1}{4k}$. Is this correct? Thank you.