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If s(t) = 3 sin(2 (t - π/6)) + 5, then the derivative is
s'(t) = 3 cos(2 (t - π/6)) • 2 = 6 cos(2 (t - π/6))
The critical points of s(t) occur at the values of t where s'(t) is zero or undefined. s'(t) is continuous everywhere, so we only need worry about the first case. We have
6 cos(2 (t - π/6)) = 0
cos(2t - π/3) = 0
2t - π/3 = arccos(0) + nπ
(where n is any integer)
2t - π/3 = π/2 + nπ
2t = 5π/6 + nπ
t = 5π/12 + nπ/2
If you're only looking for t in the interval [0,2π), then you have four critical points at t = 7π/12, t = 11π/12, t = 17π/12, and t = 23π/12.