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
1. $f(x)=2sin(3x) \Longrightarrow y=2sin(3x) \Longrightarrow x=2sin(3y) \Longrightarrow y=\frac{1}{3}sin^{-1}(\frac{x}{2}) \Longrightarrow f^{-1}(x)=\frac{1}{3}sin^{-1}(\frac{x}{2})$
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 $[-\frac{\pi}{6},\frac{\pi}{6}]$
