For what values of k, ceil ((6k+1) pi /6) =2+floor ((6k-1) pi/6) where k is a positive integer greater than equal to one?
It is possible for every integer.
The equation is this whose graph is drawn ceil ((6k+1) pi /6) – 2 – floor ((6k-1) pi/6) = 0
@Faizan Ahmed -It is not possible for k=39, 46 and many more.