Answer
There are two horizontal tangents over the specified interval, one at the point $(\frac{\pi}{3}, \frac{\pi\sqrt 3 +3}{3})$ and the second at $(\frac{2\pi}{3}, \frac{2\pi\sqrt 3 -3}{3})$.
Work Step by Step
Using the power rule $((x^n)'=nx^{n-1})$ and remembering that $((\frac{d}{dx}cos(x)=-sin(x))$ we get that $y'=\sqrt 3-2sin(x)$. Horizontal tangent$\rightarrow y'=0 \rightarrow \sqrt 3 -2sin(x)=0\rightarrow sin(x)= \frac{\sqrt 3}{2}$ $\rightarrow x=\frac{\pi}{3}$ or $x=\frac{2\pi}{3}$ (The solution of this sine function is the angle and its supplement). To find the y-coordinate substitute the x-values back into the original equation to get both points $(\frac{\pi}{3}, \frac{\pi\sqrt 3 +3}{3})$ and $(\frac{2\pi}{3}, \frac{2\pi\sqrt 3 -3}{3})$.
