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I guess you mean a polynomial $p(x)$ with rational coefficients. Then, once $\cos {\pi/n}$ is a root of $p(x)$, $\deg p=d$, $e^{i\pi/n}$ is a root of a polynomial $t^dp((t+1/t)/2)$. But $e^{i\pi/n}$ is a root of a cyclotomic polynomial $g(t)=\Phi_{2n}(t)$, which is irreducible, thus $\Phi_{2n}(t)$ should divide $t^dp((t+1/t)/2)$, that is, for any $k\in \{0,1,\dots,2n-1\}$ coprime to $2n$ the number $\cos \pi k/n$ is also a root of $p(x)$. For $k$ and $2n-k$ we get the same value of a cosine, so we get $\varphi(2n)/2$ different roots. Actually the polynomial with all these $\varphi(2n)/2$ roots has rational coefficients. To see this observe that $\Phi_{2n}(t)=t^{\varphi(2n)/2}H(t+1/t)$ for some polynomial $H$, which of course has rational (even integer) coefficients, and this $H$ has roots $2\cos \pi k/n$ for $k$ coprime to $2n$.