For full details of this and more, the best place to look is the
following paper:
D. S. Dummit, Solving solvable quintics. Math. Comp. 57 (1991), 387-401.
The main idea (which extends to any equation with a cyclic Galois group)
is to consider Lagrange resolvents. Let the equation have roots $x_1,\ldots,x_5$
with an element $\tau$ of the Galois group permuting them as $x_i\mapsto x_{i+1}$.
Let $\zeta=\exp(2\pi i/5)$ be the standard fifth root of unity. Then the Lagrange resolvents are
$\begin{align*}
A_0&=x_1+x_2+x_3+x_4+x_5\\
A_1&=x_1+\zeta x_2+\zeta^2 x_3+\zeta^3 x_4+\zeta^4 x_5\\
A_2&=x_1+\zeta^2 x_2+\zeta^4 x_3+\zeta x_4+\zeta^3 x_5\\
A_3&=x_1+\zeta^3 x_2+\zeta x_3+\zeta^4 x_4+\zeta^2 x_5\\
A_4&=x_1+\zeta^4 x_2+\zeta^3 x_3+\zeta^2 x_4+\zeta x_5
\end{align*}$
Once one has $A_0,\ldots,A_4$ one easily gets $x_1,\ldots,x_5$.
It's easy to find $A_0$ :-)
The point is that $\tau$ takes $A_j$ to $\zeta^{-j}A_j$
and so takes $A_j^5$ to $A_j^5$. Thus $A_j^5$ can be written down in terms
of rationals (if that's your starting field) and powers of $\zeta$.
Alas, here is where the algebra becomes difficult. The coefficients
of powers of $\zeta$ in $A_1^5$ are complicated. They can be expressed
in terms of a root of a "resolvent polynomial" which will have a rational
root as the equation is cyclic. Once one has done this, you have $A_1$
as a fifth root of a certain explicit complex number. Then one can
express the other $A_j$ in terms of $A_1$. The details are not very pleasant,
but Dummit skilfully navigates through the complexities, and produces
formulas which are not as complicated as they might be. Alas, I don't have
the time nor the energy to provide more details.
Maybe off topic. For polynomial:
$${x}^{5}+q\,{x}^{4}+p\,{x}^{3}+t\,{x}^{2}+k\,x-m=0$$
with root R, characteristic polinomial of matrix:
$$\small\begin{pmatrix}a & -c\,t-b\,p+g\,m-d\,k & -b\,t+d\,m-c\,k & c\,m-b\,k & b\,m\cr b & b\,q+a & -c\,t+g\,m-d\,k & d\,m-c\,k & c\,m\cr c & c\,q+b & b\,q+c\,p+a & g\,m-d\,k & d\,m\cr d & d\,q+c & c\,q+d\,p+b & d\,t+b\,q+c\,p+a & g\,m\cr g & g\,q+d & d\,q+g\,p+c & g\,t+c\,q+d\,p+b & d\,t+b\,q+c\,p+g\,k+a\end{pmatrix}$$
can have a root:
$$\small{x=g\,{R}^{4}+\left( g\,q+d\right) \,{R}^{3}+\left( d\,q+g\,p+c\right) \,{R}^{2}+\left( g\,t+c\,q+d\,p+b\right) \,R+d\,t+b\,q+c\,p+g\,k+a}$$
For example:$${x}^{5}-m=0$$
characteristic polinomial of matrix:
$$\begin{pmatrix}a & g\,m & d\,m & c\,m & b\,m\cr b & a & g\,m & d\,m & c\,m\cr c & b & a & g\,m & d\,m\cr d & c & b & a & g\,m\cr g & d & c & b & a\end{pmatrix}$$
has a root:
$$x=g\,{m}^{\frac{4}{5}}+d\,{m}^{\frac{3}{5}}+c\,{m}^{\frac{2}{5}}+b\,{m}^{\frac{1}{5}}+a$$
For example:
$${x}^{5}+{x}^{4}-4\,{x}^{3}-3\,{x}^{2}+3\,x+1=0$$
has a root:
$$x=2\,\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right)$$
if a=1,b=1,c=1,d=1,g=1
characteristic polinomial
$$-{x}^{5}-6\,{x}^{4}+{x}^{3}+10\,{x}^{2}+6\,x+1$$
has a root:
$$x=16\,{\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) }^{4}+16\,{\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) }^{3}-8\,{\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) }^{2}-10\,\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) -2$$
Sorry for my bad English.
Best Answer
Let $2\cos{x}=t$.
Thus, we need to solve $$t^5-4t^3-t^2+11t-7=0$$ or $$t^5-t^4+t^4-t^3-3t^3+3t^2-4t^2+4t+7t-7=0$$ or $$(t-1)(t^4+t^3-3t^2-4t+7)=0$$ and since by AM-GM $$t^4+t^3-3t^2-4t+7=(t^4+3-4t)+\left(2\cdot\frac{t^3}{2}+4-3t^2\right)\geq$$ $$\geq4\sqrt[4]{t^4\cdot1^3}-4t+3\sqrt[3]{\left(\frac{t^3}{2}\right)^2\cdot4}-3t^2=4(|t|-t)>0,$$ we obtain $t=1$ and $\cos{x}=\frac{1}{2}$ only.