Simplify $\arccos\left(2\cos x\right)$. Let $x\in[\pi/3,2\pi/3]$. We know that $\arccos (\cos x)=x$ but what we can say about $\arccos\left(2\cos x\right)$?

Are there, for example, any "half-angle formula" also for inverse trigonometric functions?
As mentioned by others already, there is no nice simplification of $\cos^{-1}(2\cos(x))$, but the following similar-looking identities hold:

$\cos(2\cos^{-1}(x)) = 2x^2-1$ for any $x \in [-1,1]$.
$\cos(\frac12\cos^{-1}(x)) = \sqrt{\frac12(x+1)}$ for any $x \in [-1,1]$.

Both can be easily proven by the double-angle formula.
$\arccos(2\cos x)$ can not be defined for $x∈(0,π)$ it can only be defined for $x∈[π/3,2π/3]$. So I don't think it is useful to make calculation for a function with such a domain (Neither useful nor practical)