Answer:
D. [tex]y=3\cos(\frac{4}{3}x-2\pi)+2[/tex]
Step-by-step explanation:
The given function is:
[tex]y=-3\sin(\frac{2}{3}x-2\pi)+2[/tex]
This function is of the form:
[tex]y=A\sin(Bx-C)+D[/tex], where [tex]B=\frac{2}{3}[/tex]
The period is given by:
[tex]T=\frac{2\pi}{B}[/tex]
[tex]T=\frac{2\pi}{\frac{2}{3}}=3\pi[/tex]
Half of this period is [tex]\frac{3\pi}{2}[/tex].
The function that has a period of [tex]\frac{3\pi}{2}[/tex] is
[tex]y=3\cos(\frac{4}{3}x-2\pi)+2[/tex]
