
Hmmmm. You are not asked for r, you are asked for period, in time.
3t= 2PI*n n=0, 1, ...
t= 2/3 PI n try n=1 for the first rev.

I was looking for r because period = [2(pi)(r)]/v......
And how did you get to 3t=2Pi*n???? What equation is that?????

solve t by the method I gave, for get r.
3t=0, 2PI, 4PI....is not the period of a sin function =2PI or a multiple thereof?


Well, just think about it logically: if the equation were x=5cos(t) and y=5sin(t), then the period of revolution (time for it to complete one cycle) would be 2 pi, because that's the period of the sin/cos function. However, since it's 3t instead of t, the cycle of rotation would be much faster: set 3t=2*pi, so t, or the new period of revolution, is 2*pi/3.

Well, I am 14 years late to answer this, but just hear me out!
We can see that the x component of the coordinate is x=2 cos(3t)
Again, recall that in a circular motion the coordinates can be written as:
A cos wt (for x coordinate) {Here w is omega, the angular velocity}
See the relation? A cos wt <=> 2 cos 3t?
omega=3
Also, omega = 2*pi*frequency
Or omega=(2*pi)/Time period
3=2pi/T
T=2/3 pi or 2*pi/3

Sorry, I meant 5 cos wt and not 2 cos wt
Please ignore that