That's what I'm trying to explain with a more simple example but maybe the link isn't clear.
You know that at times a and b, if b - a = 2pi, that is, the difference between the two values is 2pi, then cos(b) = cos(a). For example, for cosx, two values are a = 0 and b = 2pi, in which case b-a satisfies the condition. This means that between a and b, there has been exactly one
period of cos, which is exactly the same as one revolution of the wheel.
The function you're dealing with is Rcos(pit/5 + c) where the R has no bearing on the length it takes for a revolution because, well, its constant. And you already know that if b - a = 2pi then there's one revolution between times a and b. What's a and b in this case? The variable will be time, and it will look like
b=0.2π×T2+c,a=0.2π×T1+c where t1 and t2 correspond to values a and b such that b - a = 2pi. So we need to find the values of t.
First, the constants cancel, so you get b - a = 0.2pi(T2 - T1) = 2pi. If this is true then T2 - T1 is the time taken for one revolution. It doesn't matter what they are as long as its true, so for simplicity set T1 = 0 and solve instead of trying to find two T1s and T2s by trial and error