You need to solve the following equation for $z=e^{ix}$
$$\left(z^2+1\right) \left(z^{2 n}-1\right)-n \left(z^2-1\right) \left(z^{2 n}+1\right)=0.$$
[Wolfram Alpha](https://www.wolframalpha.com/input?i=Solve%5B-%28%281+%2B+z%5E2%29+%28-1+%2B+z%5E%2814%29%29%29+%2B++++++7+%28-1+%2B+z%5E2%29+%281+%2B+z%5E%2814%29%29+%3D%3D+0%2C+z%5D) can do that for you. The answer is in radicals for $n\leq 7$,for example
$$n=7\Rightarrow z=\sqrt{\frac{1}{6} \left(\sqrt{7}+i \sqrt{2 \left(\sqrt{7}+14\right)}-1\right)}.$$
Note that $|z|=1$, so $x$ is real. This is one of the solutions, for the others you vary the sign of the square roots (and there are also roots at $\pm 1$ and $\pm i$, see figure).
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