Trigonometry in this calculus problem

[endeavor]

Find all the points on the graph of the function f(x) = 2 \sin x + \sin^2 x at which the tangent is horizontal.
f'(x) = 2 \cos x + 2 \sin x \cos x = 0
2 \cos x (1 + \sin x) = 0
\cos x = 0 or \sin x = -1
x = \frac{\pi}{2} + n\pi or x = \frac{3\pi}{2} + 2n\pi
I then plug into find the y-coordinates. However, the book's answer for the cos x = 0 part is x = \frac{\pi}{2} + 2n\pi. But cos x = 0 for all x = \frac{\pi}{2} + n\pi... right? where did the 2 come from?

 

[courtrigrad]

f'(x) = 2\cos x + 4\sin x \cos x

2\cos x(1+2\sin x) = 0

 

[endeavor]

oops. It's actually f(x) = 2 \sin x + \sin^2 x