Hitago

@CPhill: Thank you, unfortunately the message was incomplete, sorry for the inconvenience. I've attached a picture of the complete message, would you like to answer to it? Thanks and sorry again.

I don't know why but I see the question incomplete, I don't know if it is just a problem of mine. I will attach a pic below:

Thank you Alan! Your solution is clear and useful. Do you agree with my reasoning of why the use of triangle inequality is incorrect to determine the maximum or I have wrongly analyzed the mistake as well?

No maybe I got it. I would be a problem if I I associate at the same triangle two different perimeters, and this is impossible; my mistake.

Thanks for your answer, I have a doubt: why for $x=0$ it is $\arccos x$ not a function? I know that I must restrict the image of the cosine, conventionally, to $[0,\pi]$ to get an injective function so I can invert it, but it doesn't seem to me that $x=0$ creates a problem...indeed $\arccos(0)=\frac{\pi}{2}$. What am I missing? :D

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.

Thank you, now the part on the signs is clear; however, I still don't get why $2\alpha \in \left(\frac{\pi}{2},\pi\right]$. For example, why not $2\alpha \in \left(0,\frac{\pi}{2}\right]$? That would change the sign of $\cos (2\alpha)$, so I don't believe that this other choice is equivalent, so I suppose that there must be a criterion to take the right choice. Can you help me again? Thanks.