TITLE: Proving $2 ( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} )$ is a root of$ \sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$
QUESTION [24 upvotes]: How can one show that the number
$2 \left( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} \right)$
is a root of the equation
$\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$?

REPLY [2 votes]: Sorry if this is a little off-topic here, but the question where it was topical was marked as a duplicate of this one.
Well, $2$ is a primitive root $\pmod{19}$, so we can group the nonzero residues $2^n$ according to the congruence class of $n\pmod{18}$.
$$\begin{array}{ccccccc}
r_0&\color{red}{1}&\color{red}{8}&\color{red}{7}&\color{red}{18}&\color{red}{11}&\color{red}{12}\\
r_1&\color{green}{2}&\color{green}{16}&\color{green}{14}&\color{green}{17}&\color{green}{3}&\color{green}{5}\\
r_2&\color{blue}{4}&\color{blue}{13}&\color{blue}{9}&\color{blue}{15}&\color{blue}{6}&\color{blue}{10}\\
\end{array}$$
From the problem statement, clearly we are interested in $r_1$. To work out the products of $r_0$, $r_1$,and $r_2$ we can write out tables were we add all the elements $\pmod{19}$. The table for $r_0\times r_1$:
$$\begin{array}{ccccccc}
 &\color{green}{2}&\color{green}{16}&\color{green}{14}&\color{green}{17}&\color{green}{3}&\color{green}{5}\\
\color{red}{1}&\color{green}{3}&\color{green}{17}&\color{blue}{15}&\color{red}{18}&\color{blue}{4}&\color{blue}{6}\\
\color{red}{8}&\color{blue}{10}&\color{green}{5}&\color{green}{3}&\color{blue}{6}&\color{red}{11}&\color{blue}{13}\\
\color{red}{7}&\color{blue}{9}&\color{blue}{4}&\color{green}{2}&\color{green}{5}&\color{blue}{10}&\color{red}{12}\\
\color{red}{18}&\color{red}{1}&\color{blue}{15}&\color{blue}{13}&\color{green}{16}&\color{green}{2}&\color{blue}{4}\\
\color{red}{11}&\color{blue}{13}&\color{red}{8}&\color{blue}{6}&\color{blue}{9}&\color{green}{14}&\color{green}{16}\\
\color{red}{12}&\color{green}{14}&\color{blue}{9}&\color{red}{7}&\color{blue}{10}&\color{blue}{15}&\color{green}{17}\\
\end{array}$$
From the above we can see that $r_0\times r_1=r_0+2r_1+3r_2$. The table for $r_1\times r_1$:
$$\begin{array}{ccccccc}
 &\color{green}{2}&\color{green}{16}&\color{green}{14}&\color{green}{17}&\color{green}{3}&\color{green}{5}\\
\color{green}{2}&\color{blue}{4}&\color{red}{18}&\color{green}{16}&\color{black}{0}&\color{green}{5}&\color{red}{7}\\
\color{green}{16}&\color{red}{18}&\color{blue}{13}&\color{red}{11}&\color{green}{14}&\color{black}{0}&\color{green}{2}\\
\color{green}{14}&\color{green}{16}&\color{red}{11}&\color{blue}{9}&\color{red}{12}&\color{green}{17}&\color{black}{0}\\
\color{green}{17}&\color{black}{0}&\color{green}{14}&\color{red}{12}&\color{blue}{15}&\color{red}{1}&\color{green}{3}\\
\color{green}{3}&\color{green}{5}&\color{black}{0}&\color{green}{17}&\color{red}{1}&\color{blue}{6}&\color{red}{8}\\
\color{green}{5}&\color{red}{7}&\color{green}{2}&\color{black}{0}&\color{green}{3}&\color{red}{8}&\color{blue}{10}\\
\end{array}$$
This says that $r_1\times r_1=6+2r_0+2r_1+r_2$. And finally the table for $r_2\times r_1$:
$$\begin{array}{ccccccc}
 &\color{green}{2}&\color{green}{16}&\color{green}{14}&\color{green}{17}&\color{green}{3}&\color{green}{5}\\
\color{blue}{4}&\color{blue}{6}&\color{red}{1}&\color{red}{18}&\color{green}{2}&\color{red}{7}&\color{blue}{9}\\
\color{blue}{13}&\color{blue}{15}&\color{blue}{10}&\color{red}{8}&\color{red}{11}&\color{green}{16}&\color{red}{18}\\
\color{blue}{9}&\color{red}{11}&\color{blue}{6}&\color{blue}{4}&\color{red}{7}&\color{red}{12}&\color{green}{14}\\
\color{blue}{15}&\color{green}{17}&\color{red}{12}&\color{blue}{10}&\color{blue}{13}&\color{red}{18}&\color{red}{1}\\
\color{blue}{6}&\color{red}{8}&\color{green}{3}&\color{red}{1}&\color{blue}{4}&\color{blue}{9}&\color{red}{11}\\
\color{blue}{10}&\color{red}{12}&\color{red}{7}&\color{green}{5}&\color{red}{8}&\color{blue}{13}&\color{blue}{15}\\
\end{array}$$
And so $r_2\times r_1=3r_0+r_1+2r_2$. We also know by symmetry that $0=1+r_0+r_1+r_2$, and we can work out 
$$\begin{align}r_1^3&=r_1\times(6+2r_0+2r_1+r_2)\\
&=6r_1+2(r_0+2r_1+3r_2)+2(6+2r_0+2r_1+r_2)+3r_0+r_1+2r_2\\
&=12+9r_0+15r_1+10r_2\end{align}$$
so we can set up the system
$$\begin{array}{ccccc}r_1^3&=12&+9r_0&+15r_1&+10r_2\\
r_1^2&=6&+2r_0&+2r_1&+r_2\\
r_1&=&&r_1&\\
0&=1&+r_0&+r_1&+r_2\end{array}$$
Eliminating $r_0$ and $r_2$, we find that
$$r_1^3+r_1^2-6r_1-7=0$$
Let $r_1=x-\frac13$
$$x^3-\frac{19}3x-\frac{133}{27}=0$$
Using Vieta's trigonometric solution we find that 
$$r_1=-\frac13+\frac23\sqrt{19}\cos\left(\frac13\cos^{-1}\left(\frac7{2\sqrt{19}}\right)+\frac{2\pi k}3\right)$$
Comparing numerical values, $k=0$ corresponds to $r_1$, $k=1$ corresponds to $r_2$, and $k=2$ corresponds to $r_0$.