## Calculus 10th Edition

Over the specified the interval, the only horizontal tangent line is at $(\pi, \pi)$.
Using the power rule $((x^n)'=nx^{n-1})$ and remembering that $(\frac{d}{dx}sin(x)=cos(x))$ we get that $y'=1+cos(x)$. Horizontal tangent line $\rightarrow y'=0 \rightarrow 1+cos(x)=0 \rightarrow x=\arccos(-1) \rightarrow x=\pi$ To find the y-coordinate, plug in the x-value into the original equation to get $y=\pi+sin(\pi)=\pi$. Hence, the only horizontal tangent is at the point $(\pi, \pi)$.