Why suddenly have pie k after removing sin

For positive angles, sin0=0, sin(pi)=0, sin(2pi)=0. Second round, sin(3pi)=0, sin(4pi)=0. No limit is given, so the number of rounds go on. Same for the negative angles...
So, in general, sin(k*pi)=0 where k can take 0, +/-1, +/-2, +/-3, ...
Question asked for positive, so take only the k that make n positive.

In secondary level, we will present the working as follows:
Basic angle =0 or pi
Hence (pi /6)(n+1)= pi, 2pi, 3pi.
n = ...

But in JC, you should use k and state k=0, +/-1...