How to rewrite $\pi - \arccos(x)$ as $2\arctan(y)$? I get the following results after solving the equation $\sqrt[4]{1 - \frac{4}{3}\cos(2x) - \sin^4(x)} = -\,\cos(x)$, :
$$
x_{1} = \pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}\\
x_{2} = \pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}
$$ 
Wolfram Alpha, here the link, instead, gives the following results:
$
x_{1} = 2\pi n - 2\arctan(\sqrt{5 + 2\sqrt{6}}) : n \in \mathbb{Z}\\
x_{2} = 2\pi n + 2\arctan(\sqrt{5 + 2\sqrt{6}}) : n \in \mathbb{Z}
$
Now, supposing that my solutions are correct, this means that there must be a relation between:
$\pi - \arccos(\frac{\sqrt{6}}{3})$ and $- 2\arctan(\sqrt{5 + 2\sqrt{6}})$
or between:
$\pi + \arccos(\frac{\sqrt{6}}{3})$ and $+ 2\arctan(\sqrt{5 + 2\sqrt{6}})$
or viceversa. But, given the solutions I have found, how can I prove that they are effectively the same as the solutions Wolfram found? Mathematically?
P.S.: I have found out, by looking at the graphs of $y_{1} = \pi - \arccos(\frac{\sqrt{6}}{3})$ and $y_{2} = 2\arctan(\sqrt{5 + 2\sqrt{6}})$ e.g., that:
$\pi - \arccos(\frac{\sqrt{6}}{3}) = 2\arctan(\sqrt{5 + 2\sqrt{6}})$
Then, of course: 
$\pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi = 2\arctan(\sqrt{5 + 2\sqrt{6}})+ 2\pi \\
\pi - \arccos(\frac{\sqrt{6}}{3}) + 4\pi = 2\arctan(\sqrt{5 + 2\sqrt{6}})+ 4\pi \\
... every\,\,360°n, n \in \mathbb{Z}$
And that:
$\pi + \arccos(\frac{\sqrt{6}}{3}) = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 2\pi\\
\pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 4\pi\\
\pi + \arccos(\frac{\sqrt{6}}{3}) + 4\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 6\pi\\
...
\pi + \arccos(\frac{\sqrt{6}}{3}) + (n - 2)\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ n\pi\\
$
So we can say that $\pi + \arccos(\frac{\sqrt{6}}{3}) + (n - 2)\pi$ differs from $-2\arctan(\sqrt{5 + 2\sqrt{6}}) + n\pi$ by just one lap ($2\pi$), otherwise they can be safely considered the same (for all integers).
So how to rewrite $\arccos$ in terms of $\arctan$?
Thanks for the attention!

We define $y = \arccos \left( \dfrac{\sqrt{6}}{3} \right)$. Since $0 < \dfrac{\sqrt{6}}{3} < 1$, we have $0 < y < \dfrac{\pi}{2}$, and $\dfrac{\pi}{2} < \pi - y < \pi$. Therefore,
\begin{equation*}
\tan \left( \frac{\pi - y}{2} \right) = \sqrt{\frac{1 - \cos (\pi - y)}{1 + \cos (\pi + y)}} = \sqrt{\frac{1 + \cos y}{1 - \cos y}} = \sqrt{\dfrac{1 + \frac{\sqrt{6}}{3}}{1 - \frac{\sqrt{6}}{3}}} = \sqrt{\frac{3 + \sqrt{6}}{3 - \sqrt{6}}} = \sqrt{5 + 2\sqrt{6}}.
\end{equation*}
Then we have
\begin{equation*}
\pi - y = 2 \arctan \left( \sqrt{5 + 2\sqrt{6}} \right),
\end{equation*}
that is
\begin{equation*}
\pi - \arccos \left( \frac{\sqrt{6}}{3} \right) = 2 \arctan \left( \sqrt{5 + 2\sqrt{6}} \right),
\end{equation*}
By the same token, we have $\pi < \pi + y < \dfrac{3}{2} \pi$. Therefore,
\begin{equation*}
\begin{split}
\tan \left( \frac{\pi + y}{2} \right) &= -\sqrt{\frac{1 - \cos (\pi - y)}{1 + \cos (\pi + y)}} = -\sqrt{\frac{1 + \cos y}{1 - \cos y}} \\
&= -\sqrt{\dfrac{1 + \frac{\sqrt{6}}{3}}{1 - \frac{\sqrt{6}}{3}}} = -\sqrt{\frac{3 + \sqrt{6}}{3 - \sqrt{6}}} = -\sqrt{5 + 2\sqrt{6}}.
\end{split}
\end{equation*}
Then we have
\begin{equation*}
\pi + y = 2 \arctan \left( -\sqrt{5 + 2\sqrt{6}} \right) = -2\arctan \left( \sqrt{5 + 2\sqrt{6}} \right),
\end{equation*}
that is
\begin{equation*}
\pi + \arccos \left( \frac{\sqrt{6}}{3} \right) = -2 \arctan \left( \sqrt{5 + 2\sqrt{6}} \right).
\end{equation*}
From this exercise, you can easily prove a general trigonometric relation between $\arccos$ and $\arctan$.
