Answer
$f^{-1}(x)=sin^{-1}(\frac{x-2}{5})$
range of $f(x)$: $[-3,7]$
domain and range of $f^{-1}(x)$: $[-3,7]$ and $[-\frac{\pi}{2}, \frac{\pi}{2}]$
Work Step by Step
Step 1. $f(x)=5sin(x)+2 \longrightarrow y=5sin(x)+2 \longrightarrow x=5sin(y)+2 \longrightarrow y=sin^{-1}(\frac{x-2}{5}) \longrightarrow f^{-1}(x)=sin^{-1}(\frac{x-2}{5})$
Step 2. We can find the domain and range of $f(x)$: $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $[-3,7]$
Step 3. We can find the domain and range of $f^{-1}(x)$: $[-3,7]$ and $[-\frac{\pi}{2}, \frac{\pi}{2}]$
