Question

How do you find the period of `y=3sin(x-2pi/3)-2`?

Answer 1

the searched period is `2pi`

Explanation:

Calculating
`sin(x-2pi/3+2pi)=`
`sin(x-2pi/3)cos(2pi)+cos(x-2/3*pi)*sin(2pi)`
since `sin(2pi)=0` and `cos(2pi)=1`
we get

`sin(x-2/3*pi+2pi)=sin(x-2/3*pi)`