algebra precalculus - How is the sin function being rewritten?

I'm working through a trigonometry book and was shown this equation being worked out. I don't understand the rules for doing a particular step:

$$\begin{align} A &= A\sin(x-vt) \\ 1 &= \sin(x-vt) \\ x-vt &= {\pi \over 2} \\ x &= {\pi \over 2}+vt \end{align}$$

How are they going from $1=\sin(x-vt)$ to $x-vt = {\pi \over 2}$? Thanks!

Answer

Whenever you confront a step like this, notice that the $\sin$ is being taken off one side. That means that the inverse of the $\sin$ function must have been used. From this reasoning, even if you don't know that $\arcsin(1) = \frac{\pi}{2}$, then you can make the logical assumption that it is, and from that you can understand the step.
EDIT:
Note that $\arcsin(\sin(x))$ does not always equal $x$. I would not use this step in trying to solve an equation, but for something like this, the strategy shown above helps.