Answer
$y=3 \ \sin (\dfrac{2}{3}x+\dfrac{2}{9})$
Work Step by Step
The general form for the sinusoidal function can be expressed as:
$y=A\sin{(\omega x-\phi)}+B ..(1)$
where $A$ is the amplitude and $B$ represents the vertical shift.
The $\omega$ can be found from the period by the formula:
$\omega=\dfrac{2\pi}{T}$
and the phase shift is $\dfrac{\phi}{\omega}$.
This means that $\phi=\omega \times \ Phase \ Shift$
We have: $A=2$, $B=0$,
$\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{3 \pi}=\dfrac{2}{3}$
$\phi=(\dfrac{2}{3})( \dfrac{-1}{3})=\dfrac{-2}{9}$
Therefore, our sinusoidal function (1) becomes: $y=3 \ \sin (\dfrac{2}{3}x+\dfrac{2}{9})$