Answer:
[tex]3\pi \rightarrow y=2\cos \dfrac{2x}{3}\\ \\\dfrac{2\pi }{3}\rightarrow y=6\sin 3x\\ \\\dfrac{\pi }{3}\rightarrow y=-3\tan 3x\\ \\10\pi \rightarrow y=-\dfrac{2}{3}\sec \dfrac{x}{5}[/tex]
Step-by-step explanation:
The period of the functions [tex]y=a\cos(bx+c)[/tex] , [tex]y=a\sin(bx+c)[/tex], [tex]y=a\sec (bx+c)[/tex] or [tex]y=a\csc(bx+c)[/tex] can be calculated as
[tex]T=\dfrac{2\pi}{b}[/tex]
The period of the functions [tex]y=a\tan(bx+c)[/tex] or [tex]y=a\cot(bx+c)[/tex] can be calculated as
[tex]T=\dfrac{\pi}{b}[/tex]
A. The period of the function [tex]y=-3\tan 3x[/tex] is
[tex]T=\dfrac{\pi}{3}[/tex]
B. The period of the function [tex]y=6\sin 3x[/tex] is
[tex]T=\dfrac{2\pi}{3}[/tex]
C. The period of the function [tex]y=-4\cot \dfrac{x}{4}[/tex] is
[tex]T=\dfrac{\pi}{\frac{1}{4}}=4\pi[/tex]
D. The period of the function [tex]y=2\cos \dfrac{2x}{3}[/tex] is
[tex]T=\dfrac{2\pi}{\frac{2}{3}}=3\pi[/tex]
E. The period of the function [tex]y=-\dfrac{2}{3}\sec \dfrac{x}{5}[/tex] is
[tex]T=\dfrac{2\pi}{\frac{1}{5}}=10\pi[/tex]
