[tex]y=2\cos(3x-2\pi)\\\\\text{The period of a cosine is}\ 2\pi.\\\\P-period\\\\2\cos(3(x+P)-2\pi)=2\cos(3x-2\pi)\qquad\text{divide both sides by 2}\\\\\cos(3x+3P-2\pi)=\cos(3x-2\pi)\iff3x+3P-2\pi=3x-2\pi+2\pi\\\\3x+3P-2\pi=3x\qquad\text{subtract 3x from both sides}\\\\3P-2\pi=0\qquad\text{add}\ 2\pi\ \text{to both sides}\\\\3P=2\pi\qquad\text{divide both sides by 3}\\\\\boxed{P=\dfrac{2\pi}{3}}\\\\Answer:\ \boxed{D.\ \dfrac{2\pi}{3}}[/tex]
