How to find the at rest position of a particle when trig functions

[TitoSmooth]

so my given: s(t)=cos(pie8*t/4)

took the derivative= velocity function

then, v(t)= -pie/4 *sin(pie*t/4)

When is the particle at rest? v(t)=0

now, 0= -pie/4 *sin(pie*t/4)


im lost here. I know it's very simple I am just over thinking. What do I do from here?

thanks

 

[TitoSmooth]

figured it out. had to remind myself that the period of sin and cosin is -2pie->2pie.

then use the formula forget what its called. but 2pie/b

2pie/pie/4 = 8. then when divide the interval into 4 sub intervals and we have 0, 2, 4, 6, 8.

remembering that sin is negative at 0, pie, 2pie. so velocity is zero at t=0,4,8

 

[jbunniii]

$$-\frac{\pi}{4} \sin \frac{\pi t}{4} = 0$$
is true if and only if $\sin(\pi t/4) = 0$. For what values of $x$ is $\sin(x) = 0$?