This is an exercise taken from Mooculus-textbook (page 17, exercise 5 to be exact).
The task given is to find an inverse for $f(x) = 18\sin(\frac{x\pi}{7})+20$ (restricting domain to $[3.5,10.5]$) and interpret $f^{-1}(20)$. I'm given that $f^{-1}(20) = 7$ but what I've come up with as an inverse does not yield anything like that.
Here is how I've tried to approach this (only tool I'm aware of for finding inverses is to swap y and x, then solve for y).
$f(x) = 18\sin(\frac{x\pi}{7})+20 \Longrightarrow y = 18\sin(\frac{x\pi}{7})+20$
$x = 18\sin(\frac{y\pi}{7})+20 \Longrightarrow x-20=18\sin(\frac{y\pi}{7}) \Longrightarrow$ $\frac{x-20}{18}= \sin(\frac{y\pi}{7}) \Longrightarrow \arcsin({\frac{x-20}{18}})=\frac{x\pi}{7} \Longrightarrow$ $7\arcsin(\frac{x-20}{18}) = y\pi \Longrightarrow \frac{7\arcsin(\frac{x-20}{18})}{\pi} = y$.
But $\frac{7\arcsin(\frac{20-20}{18})}{\pi} = 0$, not 7.
The original problem in the book is:
In much the same way that $\sin(x)=0$ has multiple values for $x$, $\arcsin(0)=x$ also has multiple values for $x$. In order to make $\arcsin(x)$ a function, we assign a specific domain. The principal domain is $-\frac\pi2\le y\le\frac\pi2$. However, this question has a different domain so $\arcsin(0)$ can take different values. I will list a few possibilities(that are positive) irrespective of the domain given. $$\arcsin(0)=0,\pi, 2\pi, 3\pi,4\pi, 5\pi...n\pi \, \text{for}\ \ n \in \mathbb{Z}$$
This can easily be seen by drawing the unit circle.
The domain $X=[3.5, 10.5]$ refers to the value of $t$ and in your case $y$. So, looking at the possibilities the only appropriate answer is $\pi$.
$$\frac{7\arcsin(0)}{\pi}=\frac{7\times0}{\pi} = 0<3.5 \implies \notin X $$ $$\frac{7\arcsin(0)}{\pi}=\frac{7\times \pi}{\pi} = 3.5<7<10.5 \implies \in X $$ $$\frac{7\arcsin(0)}{\pi}=\frac{7\times 2\pi}{\pi} = 14>10.5 \implies \notin X $$
$h^{-1}(20)$ can be interpreted as the time needed to reach a height of $20$ m.