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!

To solve the equation $1 = \sin(x-vt)$, you must ask yourself: what angle $\theta$ is such that $\sin\theta = 1$? One such angle is $\frac{\pi}{2}$. (The others are $\frac{\pi}{2} \pm 2n\pi$).
