I want to calculate the point of maxima of the functions
$$f_n(x)=\frac{1}{n+1}\left(\frac{\sin\left((n+1)\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2$$
lying in the interval $(0,\pi)$. For any $n$ maxima of $f_n(x)$ can be described by the equation
$$\tan\left((n+1)\frac{x}{2}\right) = (n+1)\tan\left(\frac{x}{2}\right).$$
This equation can also be transformed to $$\sum_{k=1}^{\left\lfloor\frac{n+1}{2}\right\rfloor}(-1)^k k {n+1\choose{2k+1}}\tan^{2k}\frac{x}{2}=0.$$
Another form of this equation is $$n\sin\left(\frac{(n+2)x}{2}\right) = (n + 2) \sin\left(\frac{nx}{2}\right).$$
The solutions lie in the interval $\left[\frac{2k\pi}{n+1}, \frac{(2k+1)\pi}{n+1}\right], k = 1, 2, 3,\dots,\left\lfloor\frac{n-1}{2}\right\rfloor.$ I need solutions in terms of $k$ and $n$. I know that a closed form is quite difficult to obtain but if someone can provide me a series expansion for $x$, it will also be very helpful.
This is not a full answer but it is too long for a comment.
We can approximate the solution using one step of Newton method using $$x_0=\frac {(4k+1)\pi}{2(n+1)}$$ corresponding to the midpoint of the interval $\left[\frac{2k\pi}{n+1}, \frac{(2k+1)\pi}{n+1}\right]$ you properly defined.
$$f(x)=n\sin \left(\frac{1}{2} (n+2) x\right) - (n + 2)\sin\left(\frac{nx}{2}\right)$$ $$f'(x)=\frac{1}{2} n (n+2) \cos \left(\frac{1}{2} (n+2) x\right)-\frac{1}{2} n (n+2) \cos \left(\frac{n x}{2}\right)$$ $$x_1=x_0-\frac{f(x_0)}{f'(x_0)}$$
For illustration purposes, I looked at $n=20$; the results are given below $$\left( \begin{array}{ccc} k & x_1 & \text{exact} \\ 1 & 0.44543 & 0.42827 \\ 2 & 0.75566 & 0.73630 \\ 3 & 1.05925 & 1.03929 \\ 4 & 1.36089 & 1.34068 \\ 5 & 1.66171 & 1.64137 \\ 6 & 1.96210 & 1.94169 \\ 7 & 2.26225 & 2.24181 \\ 8 & 2.56226 & 2.54179 \\ 9 & 2.86219 & 2.84171 \end{array} \right)$$ We could do better using Halley's method instead of Newtoon. For the same example the results would be $$\left( \begin{array}{ccc} k & x_1 & \text{exact} \\ 1 & 0.42261 & 0.42827 \\ 2 & 0.72847 & 0.73630 \\ 3 & 1.03041 & 1.03929 \\ 4 & 1.33117 & 1.34068 \\ 5 & 1.63140 & 1.64137 \\ 6 & 1.93137 & 1.94169 \\ 7 & 2.23120 & 2.24181 \\ 8 & 2.53093 & 2.54179 \\ 9 & 2.83061 & 2.84171 \end{array} \right)$$
Even if the formulae are quite long, they are fully explicit in terms of $k$ and $n$.