When is a Particle at Rest in a Motion Problem?

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Sage Hopkins
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(1.)

I have a "particle in motion" problem that is asking me when a particle is at rest, which I understand to be when velocity = v(t) = 0, so

v(t) = - (π/4) sin (πt/4) = 0.

The given answer is as follows:

- (π/4) sin (πt/4) = 0

sin (πt/4) = 0

πt/4 = πn.

t = 0,4,8 seconds.

(2.) Can someone please explain to me how 0 becomes πn, and/or what specific mathematical concept(s) I need to review?
 
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What's the full problem statement?

Of course [itex]\sin(n \pi)=0[/itex] for all [itex]n \in \mathbb{Z}[/itex].
 
The full problem statement goes:

A particle moves according to a law of motion s = cos(πt/4), t >= 0, where t is measured in seconds and s in feet.

There are several sub-questions from here about velocity, acceleration, graphs, etc.., but the one that I got stuck on is

(c.) When is the particle at rest for t <= 10.

After differentiating the given function for f', understanding the answer to this question was as simple as reviewing the unit circle and the graph of sin for me, as elementary as it may be.

Thank you for your time!