How to solve $c \arccos(x)=x$ for the $x$? The equation 
$$c \arccos(x)=x$$
certainly has at least one solution for every real number in range $[-1,1]$, because the $[-1,1]$ is the domain of the $\arccos$. For example, if the $x=0.5$, then $c=0.5/\arccos(0.5) \approx 0.4775$.
The equation might be transformed to
$$
\arccos(x)=x/c \\
\cos(\arccos(x))=\cos(x/c) \\
x=\cos(x/c)
$$
but now what? 
The goal is to solve the equation for the $x$, not the $c$.
Thank You.

One cannot solve this algebraically, but you can solve it numerically. For example, one could implement a recursive approach:
$$x_0=0.9\\x_{n+1}=\cos(x_n/c)\\x=\lim_{n\to\infty}x_n$$
(The $0.9$ can actually be replaced with an real number, however, the root is generally near $0.9$)
For an example of this, see here: https://repl.it/LHoa/8
One might notice that $\cos(t/c)$ has a negative slope at $t=x$, where $x$ is the solution you seek. Thus, one could accelerate the convergence in an Euler-sum manner:
$$x_0=0.9\\x_{n+1}=\cos(x_n/c)\\a(n,0)=x_n\\a(n,k+1)=\frac{a(n,k)+a(n+1,k)}2\\x=\lim_{n\to\infty}a(n,k)=\lim_{k\to\infty}a(n,k)$$
For an example of this, see here: https://repl.it/LHoa/7
Since this is a recursive formula, we can further improve the convergence by looping initial values through: https://repl.it/LHoa/10
