Answer
$y=3\sin(\pi x).$
Work Step by Step
If a function has form $y=A\sin(\omega x)$, then there $|A|$ stands for the amplitude and $T =\frac{2\pi}{\omega}$ stands for the period.
First we identify $\omega$ and $A$ and then plug them into the form of $y=A\sin(\omega x)$.
Hence $A=|3|=3$, $\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi.$
Hence $y=A\sin(\omega x)=3\sin(\pi x).$
