How to solve for t in x'(t) = 2√3 cos(3t + 4π/3) with x' = 3

Im working a differential equation and have arrived at the form:
x'(t) = 3 = 2√3 cos(3t + 4π/3)

I need to find t when x' = 3.

Per wolfram alpha (http://www.wolframalpha.com/input/?i=3+%… it is:
t = 2πn/3 + π/6 and t = 2πn/3 - 7π/18 where n = 1, 2, 3, ....

However, in my book it shows that the answer is:
t = 2nπ/3 - π/2 and t = 2nπ/3 - 7π/18 where n = 1, 2, 3, ....

So the first one is different but the second one is the same. I trust wolfram alpha more than the book's answer but regardless, I'm not sure how to solve for t. Could you please help????

Thanks!

3 = 2√3 cos(3t + 4π/3)
√3/2= cos(3t + 4π/3)
3t + 4π/3 = +/- π/6+2nπ
3t = - 4π/3 +/- π/6+2nπ
t = - 4π/9 +/- π/18+(2/3)nπ
t = -π/2+(2/3)nπ or
t = - 7π/18 +(2/3)nπ

P.S.
You may rewrite :
t = -π/2+(2/3)nπ like
t = -π/2+(2/3)π +(2/3)(n-1)π = π/6 + (2/3)(n-1)π then result from Wolfram is right !