Trigonometry in this calculus problem

1. Oct 18, 2006

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 in to 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????

Last edited: Oct 18, 2006
2. Oct 18, 2006

$$f'(x) = 2\cos x + 4\sin x \cos x$$
$$2\cos x(1+2\sin x) = 0$$
oops. It's actually $$f(x) = 2 \sin x + \sin^2 x$$