Answer
$f^{-1}(x)=\frac{1}{3}cos^{-1}(\frac{x}{2})-\frac{2}{3}$
range of $f(x)$: $[-2,2]$,
domain and range of $f^{-1}(x)$: $[-2,2]$ and $[-\frac{2}{3},-\frac{2}{3}+\frac{\pi}{3}]$.
Work Step by Step
1. $f(x)=2cos(3x+2) \Longrightarrow y=2cos(3x+2) \Longrightarrow x=2cos(3y+2) \Longrightarrow y=\frac{1}{3}cos^{-1}(\frac{x}{2})-\frac{2}{3} \Longrightarrow f^{-1}(x)=\frac{1}{3}cos^{-1}(\frac{x}{2})-\frac{2}{3}$
2. We can find the range of $f(x)$: $[-2,2]$,
3. We can find the domain and range of $f^{-1}(x)$: $[-2,2]$ and $[-\frac{2}{3},-\frac{2}{3}+\frac{\pi}{3}]$.