The given function is
[tex]y=3\sin (\frac{1}{2}x+\frac{\pi}{6})[/tex]We have to use the following form
[tex]a\sin (bx-c)+d[/tex]Where the amplitude is a, the period is 2pi/b, and the phase shift is c/b. In the given function a = 3, b = 1/2, and c = pi/6.
[tex]\begin{gathered} a=3 \\ T=\frac{2\pi}{b}=\frac{2\pi}{\frac{1}{2}}=4\pi \\ \theta=\frac{c}{b}=\frac{\frac{\pi}{6}}{\frac{1}{2}}=\frac{2\pi}{6}=\frac{\pi}{3} \end{gathered}[/tex]