Answer
$f^{-1}(x)=\frac{1}{3}cos^{-1}(-\frac{x}{2}) $
range of $f(x)$: $[-2,2]$
domain and range of $f^{-1}(x)$: $[-2,2]$ and $[0, \frac{\pi}{3}]$
Work Step by Step
Step 1. $f(x)=-2cos(3x) \longrightarrow y=-2cos(3x) \longrightarrow x=-2cos(3y) \longrightarrow y=\frac{1}{3}cos^{-1}(-\frac{x}{2}) \longrightarrow f^{-1}(x)=\frac{1}{3}cos^{-1}(-\frac{x}{2}) $
Step 2. We can find the domain and range of $f(x)$: $[0, \frac{\pi}{3}]$ and $[-2,2]$
Step 3. We can find the domain and range of $f^{-1}(x)$: $[-2,2]$ and $[0, \frac{\pi}{3}]$
