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You need to solve the following equation for $z=e^{ix}$ $$\left(z^2+1\right) \left(z^{2 n}-1\right)-n \left(z^2-1\right) \left(z^{2 n}+1\right)=0.$$ Mathematica can do that for you, or even Wolfram Alpha. 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. None of this is particularly insightful.

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 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).

You need to solve the following equation for $z=e^{ix}$ $$\left(z^2+1\right) \left(z^{2 n}-1\right)+n \left(z^2-1\right) \left(z^{2 n}+1\right)=0.$$ Mathematica can do that for you, or even Wolfram Alpha. 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. None of this is particularly insightful.

You need to solve the following equation for $z=e^{ix}$ $$\left(z^2+1\right) \left(z^{2 n}-1\right)-n \left(z^2-1\right) \left(z^{2 n}+1\right)=0.$$ Mathematica can do that for you, or even Wolfram Alpha. 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. None of this is particularly insightful.

You need to solve the following equation for $z=e^{ix}$ $$\left(z^2+1\right) \left(z^{2 n}-1\right)+n \left(z^2-1\right) \left(z^{2 n}+1\right)=0.$$ Mathematica can do that for you, or even Wolfram Alpha. The answer is in radicals for $n\leq 7$,for example $$n=7\Rightarrow z=\sqrt{\frac{1}{6} \left(\sqrt{7}+i \sqrt{2 \left(\sqrt{7}+14\right)}-1\right)}.$$ This is one of the solutions, for the others you vary the sign of the square roots. None of this is particularly insightful.

You need to solve the following equation for $z=e^{ix}$ $$\left(z^2+1\right) \left(z^{2 n}-1\right)+n \left(z^2-1\right) \left(z^{2 n}+1\right)=0.$$ Mathematica can do that for you, or even Wolfram Alpha. 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. None of this is particularly insightful.