Consider the polynomial $x^n-a\in \mathbb Q[x]$, with $a>0$. One root of this polynomial is $\alpha$, the $n$-th root of $a$. Now let $\omega\in \mathbb C$ be a primitive $n$-th root of unity. That means that $\omega ^n=1$ and $\omega^k\ne 1$ for all $0<k<n$. So, the (typically complex) numbers $\beta_i=\omega ^i\alpha $, where $0\le i < n$, are distinct and all satisfy: $\beta _i^n=(\omega ^i\alpha )^n=(\omega ^n)^i\alpha ^n=a$. So, this shows that the polynomial has the $n$ distinct roots $\beta _0,\cdots ,\beta _n$.
(Partly) Conversely, if a polynomial $p(x)\in \mathbb Q[x]$ is of degree $n$, and $\alpha$ is a root of $p(x)$ such that for some primitive $n$-th root of unity $\omega $, it holds that $\omega^i\alpha$ is a root of $p(x)$ for all $i$, then $p(x)$ splits as $(x-\alpha)(x-\omega \alpha)(x-\omega ^2 \alpha)\cdots (x-\omega ^{n-1}\alpha)$ which is the splitting of the polynomial $x^n-(\alpha^n)$. Writing $a=\alpha ^n$ this shows that $p(x)=x^n-a$.
Galois theory provides some machinery for this:
Suppose $\rm L/K$ is Galois with $\rm G=Gal(L/K)$ and $\rm m(x):=minpoly_{\alpha,K}(x)$. Then
$\quad \rm m(\sigma(\alpha))=\sigma(m(\alpha))=\sigma(0)=0$ implies $\rm (x-\sigma\alpha)\mid m$ in $\rm L[x]$ for all $\rm \sigma\in G$,
$\quad \rm(x-\sigma\alpha)$ all coprime, $\rm \sigma\in G/S$, $\rm S=Stab_G(\alpha)$, implies $\rm f(x):=\prod\limits_{\sigma\in G/S}(x-\sigma\alpha)\mid m$ in $\rm L[x]$,
$\quad \rm \sigma f(x)=f(x)$ for all $\sigma\in G$ implies $\rm f(x)\in K[x]$; $\rm f(\alpha)=0$ implies $\rm m(x)\mid f(x)$ in $\rm K[x]$,
$\quad \rm f(x)\mid m(x)$ and $\rm m(x)\mid f(x)$ and both $\rm f,m$ monic implies $\rm f(x)=m(x)$.
Therefore the zeros of $\rm\alpha$'s minimal polynomial over $\rm K$ are precisely its $\rm Gal(L/K)$-conjugates.
As ${\rm Gal}\big({\bf Q}(\zeta_n)/{\bf Q}\big)=({\bf Z}/n{\bf Z})^\times$ it suffices to consider $\sigma:\zeta\mapsto\zeta^k$ for $k=1,\cdots,12$.
By symmetry we need only consider $1,\cdots,6$ for $\alpha=\zeta+\zeta^{-1}$. Thus
$$\begin{array}{ll} {\rm minpoly}_{\zeta+\zeta^{-1},\bf Q}(x) = & ~~~~\left(x-(\zeta+\zeta^{-1})\right)\left(x-(\zeta^2+\zeta^{-2})\right)\left(x-(\zeta^3+\zeta^{-3})\right) \\ & \times\left(x-(\zeta^4+\zeta^{-4})\right)\left(x-(\zeta^5+\zeta^{-5})\right)\left(x-(\zeta^6+\zeta^{-6})\right). \end{array}$$
Simplify the resulting expansion via the rules $\zeta^n=\zeta^{n\,\bmod\,13}$ and $\sum\limits_{k=0}^{12}\zeta^k=0$.
Best Answer
Because the sextic polynomial is palindromic, finding roots can be reduced to finding roots of the corresponding cubic in $t+t^{-1}$. A few computations show that for $u=t+t^{-1}$ we have \begin{eqnarray*} u^3&=&(t+t^{-1})^3=t^3+3t+3t^{-1}+t^{-3}\\ u^2&=&(t+t^{-1})^2=t^2+2+t^{-2}, \end{eqnarray*} which shows that \begin{eqnarray*} t^6-t^5+t^4-t^3+t^2-t+1 &=&t^3\Big((t+t^{-1})^3-(t+t^{-1})^2-2(t+t^{-1})+1\Big)\\ &=&t^3(u^3-u^2-2u+1). \end{eqnarray*} Note that for every positive integer $n$ you have $$\zeta_n+\zeta_n^{-1}=2\cos(2\pi/n),$$ which immediately shows that $2\cos(2\pi/14)$ is a root of the cubic, because $\zeta_{14}$ is a root of the sextic. Reversing the argument above also shows that if this cubic is not the minimal polynomial of $2\cos(2\pi/14)$, then the sextic is not the minimal polynomial of $\zeta_{14}$.
Now Cardano's formula yields explicit expressions for the roots $\alpha_1$, $\alpha_2$ and $\alpha_3$ of the cubic $$u^3-u^2-2u+1=0,$$ and then solving $t+t^{-1}=\alpha_k$, which simplify to quadratic polynomials in $t$, yields the roots of the sextic. I haven't done the computations myself, but I assume that plugging the coefficients of the cubic above into Cardano's formula yields $$u=\frac13\left(1-\omega^kC-\frac7{\omega^kC}\right),\qquad k\in\{0,1,2\},$$ where $C=\sqrt[3]{\frac{7+7\sqrt{27}i}2}$.