The function $y=\cos x+x$ passes the horizontal lines test, so it should have an inverse. But a quick Google search tells you that while such a function does exist it cannot be defined in elementary terms.
Is it possible to approximate an inverse function? Is doing so ever practical?
You can use Numerical Analysis to invert by finding the solutions of $y = x + \cos x$ using Bisection or Newton's Method.
Another approach is when you need an approximation locally, fit a polynomial there or something like a spline.
There are several methods like bisection method, secant method, Muller's method, Newton's method, false position method.
as @gt6989b mentioned if you need an approximation locally then you can use some polynomial interpolations like polynomial interpolation in the Lagrange form or in the Newton form.
In the case of your example $f(x):=x+\cos x$ you can make use of the periodicity of $\cos$. Note that $$f(x+2\pi)=f(x)+2\pi\qquad(-\infty<x<\infty)\ .\tag{1}$$ Assume that you have worked out a decent approximation $g$ of $f^{-1}$ in the interval $[f(0),2\pi]$: $$g:\quad \bigl[f(0),f(2\pi)]\to[0,2\pi],\quad y\mapsto g(y)\ ,$$ say a polynomial in $y$. This means that $$g: \quad[1,2\pi +1]\to[0, 2\pi]\ ,$$ and $$g\bigl(f(x)\bigr)\approx x\qquad(0\leq x\leq 2\pi)\ .$$ You are satisfied with the precision; in particular $g\bigl(1\bigr)=0$ and $g(2\pi+1)=2\pi$.
For an arbitrary $y\in{\mathbb R}$ let $k_y:=\bigl\lfloor{y\over2\pi}\bigr\rfloor$. As a consequence of $(1)$ you then have the approximation $$f^{-1}(y)\approx g(y-2\pi k_y)+2\pi k_y\qquad(-\infty<y<\infty)\ .$$
