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