The Domain of $f(x)$ is $(0,1)$.
When finding the domain of $f(\ln(x))$ We can say: $0<\ln x<1$ and apply the inverse $\ln$ function $\exp$.
We get $e^0 < e^{\ln(x)} < e^1$ which leads to $1<x<e$.
When trying to apply the same method on $f(\cos(x))$ we get : $\cos^{-1}0 < x < \cos^{-1}1$ which leads to $\frac{\pi}{2} < x < 0$ .
I know that this is flawed because the domain of $f(\cos(x))$ while : $0 < \cos(x) < 1$ should be: $-\frac{\pi}{2} < x < \frac{\pi}{2}$ (neglecting periodicity).
Where is the flaw in this method for the use on $\cos$ function?
$\cos$ is not bijective. When you use the inverse cosine function, you implicitly restrict your study on a half-period, for example on $[0, \pi]$ (if you want to use the classic definition of $\arccos$ or $\cos^{-1}$), where $\cos$ is strictly decreasing. Note that $\arccos$ is also a decreasing function therefore you have to reverse the order of the inequality.
$ 0 < \cos(x) < 1 \Leftrightarrow \arccos(0) > x > \arccos(1) $
And $\arccos(0)=\frac\pi2$ and $\arccos(1)=0$
Which gives you the final inequality $\frac\pi2 > x > 0.$
And this is expected, since these are truly the only values of $x$ that satisfty $0< \cos(x) < 1$ when $x$ belongs to $[0,\pi]$.
If you want to extend the result to all real values, then you have to use the symmetry ($\cos$ is even) to see that the answer is $(-\frac\pi2,0)\cup(0,\frac\pi2)$ on $[-\pi,\pi]$ and then use the periodicity to see that every interval $(2k\pi-\frac\pi2, 2k\pi)\cup(2k\pi, 2k\pi+\frac\pi2)$ works for $k$ integer.