Determine the period length of
y= 2/3 sin (3x/2 )
I had 3pi/2 but the correct answer is 4pi/3 how do you get that???
period of sin kx is 2π/3, (see your other post)
so for your function,
period = 2π/(3/2)
= 2π(2/3) = 4π/3
y= 2/3 sin (3x/2 )
I had 3pi/2 but the correct answer is 4pi/3 how do you get that???
so for your function,
period = 2π/(3/2)
= 2π(2/3) = 4π/3