In the same vein as Max Alekseyev's answer: if $q = \exp(i \pi/n)$ and $z = -\cos(\pi/n) \pm \sqrt{\cos(2\pi/n)}$, we have ${q}^{4}+4\,{q}^{3}z-4\,{q}^{2}{z}^{2}-2\,{q}^{2}+4\,qz+1=0$. $q$ is a root of the cyclotomic polynomial $C_{2n}$, so $z$ is a root of the resultant of ${Q}^{4}+4\,{Q}^{3}Z-4\,{Q}^{2}{Z}^{2}-2\,{Q}^{2}+4\,QZ+1$ and $C_{2n}(Q)$ with respect to $Q$. For at least the first $200$ integers $n \ge 2$ (and I suspect for all of them), this resultant is of the form $p(Z)^2$ where $p$ is irreducible over the rationals (and thus $p(Z)$ is the minimal polynomial of $z$).

In the same vein as Max Alekseyev's answer: if $q = \exp(i \pi/n)$ and $z = -\cos(\pi/n) \pm \sqrt{\cos(2\pi/n)}$, we have ${q}^{4}-4\,{q}^{3}z-4\,{q}^{2}{z}^{2}-2\,{q}^{2}-4\,qz+1=0$. $q$ is a root of the cyclotomic polynomial $C_{2n}$, so $z$ is a root of the resultant of ${Q}^{4}-4\,{Q}^{3}Z-4\,{Q}^{2}{Z}^{2}-2\,{Q}^{2}-4\,QZ+1$ and $C_{2n}(Q)$ with respect to $Q$. For at least the first $200$ integers $n \ge 2$ (and I suspect for all of them), this resultant is of the form $p(Z)^2$ where $p$ is irreducible over the rationals (and thus $p(Z)$ is the minimal polynomial of $z$).