$\arctan{x}=\arccos{2x}.$ First, let's determine some domains. By definition, it follows that $\arccos:[-1,1]\rightarrow[0,\pi]$ and $\arctan:\mathbb{R}\rightarrow\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Since the domain of $\arctan$ is the entire reals, the restricting factor here should be the domain of arccos. This means that $2x\in[-1,1]\Rightarrow x\in\left[-\frac{1}{2},\frac{1}{2}\right]$. 
I now take cosine of both sides and get $$\cos{(\arctan{x})}=\cos{(\arccos{2x})}=2x.$$
I use the formular $\cos{a}=\frac{1}{\sqrt{1+\tan^2{a}}},$ and rewrite as
$$\frac{1}{\sqrt{1+\tan^2{(\arctan{x})}}}=\frac{1}{\sqrt{1+x^2}}=2x.$$
Squaring both sides, simplifying and setting $t=x^2$ gives the quadratic equation
$$t^2+t-\frac{1}{4}=0\Longleftrightarrow \left\{
  \begin{array}{rcr}
    t_1 & = & -\frac{1}{2}-\frac{\sqrt{2}}{2} <0 \\
    t_2 & = & -\frac{1}{2}+\frac{\sqrt{2}}{2} >0\\
  \end{array}
\right.$$
Since $t_1<0$, it is a false root since a square of a real number $x$ can not become a negative real number $t$. So the final two roots in terms of $x$ are
$$\left\{\begin{array}{rcr}
    x_1 & = & +\sqrt{-\frac{1}{2}+\frac{\sqrt{2}}{2}} = \sqrt{\frac{1}{2}(\sqrt{2}-1)} \\
    x_2 & = & -\sqrt{-\frac{1}{2}+\frac{\sqrt{2}}{2}} = -\sqrt{\frac{1}{2}(\sqrt{2}-1)} \\
\end{array}
\right.$$
Clearly, both $x_1,x_2\in[-\frac{1}{2},\frac{1}{2}]$. Only $x_1$ is correct however. I did something wrong setting up the domain didin't I? I have a feeling that the correct domain for the solutions is $[0,\frac{1}{2}],$ I don't know how to see it.

Since $x \in \left[\frac{1}{2}, \frac{1}{2}\right]$ we have $\arctan x \in \left[-\arctan\frac{1}{2}, \arctan\frac{1}{2}\right] \subseteq \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $\cos(\arctan x) \ge 0$.
So $\cos(\arctan x) = 2x$ implies $x \geq 0$, which discards the negative solution $x_2$.
