Was hoping someone could help me find the inverse of $f(x) = 3x + \cos(x)$
The steps I took were:
$y = 3x + \cos(x)$
$x = 3y + \cos(y)$
$x - 3y = \cos(y)$
$\arccos(x-3y) = y $
But I still have a $y$ in the $\arccos$ , and I don't know how to isolate it.
$$y=3x+\cos x$$ $$y-3x=\cos x$$ $$x=\cos^{-1}(y-3x)$$ when $|y-3x|<1$ $$x=\cos^{-1}(y-3\cos^{-1}(y-3\cos^{-1}(y-3\cos^{-1}(y-3\cos^{-1}(y-3\cos^{-1}(y-3(....)))))$$