Answer
$f^{-1}(x)=\frac{1}{3}sin^{-1}(\frac{x}{2})$,
range of $f(x)$: $[-2,2]$
domain and range of $f^{-1}(x)$: $[-2,2]$ and $[-\frac{\pi}{6}, \frac{\pi}{6}]$
Work Step by Step
Step 1. $f(x)=2sin(3x) \longrightarrow y=2sin(3x) \longrightarrow x=2sin(3y) \longrightarrow y=\frac{1}{3}sin^{-1}(\frac{x}{2}) \longrightarrow f^{-1}(x)=\frac{1}{3}sin^{-1}(\frac{x}{2})$
Step 2. We can find the domain and range of $f(x)$: $[-\frac{\pi}{6}, \frac{\pi}{6}]$ and $[-2,2]$
Step 3. We can find the domain and range of $f^{-1}(x)$: $[-2,2]$ and $[-\frac{\pi}{6}, \frac{\pi}{6}]$
