## Thomas' Calculus 13th Edition

Yes, $x=\pi$
Step 1. Given $y=x+sin(x)$, we have $y'=1+cos(x)$ Step 2. For a horizontal tangent, let $y'=0$, we have $cos(x)=-1$ and $x=\pi$ in $[0,2\pi]$ Step 3. At $x=\pi$, $y=\pi+sin(\pi)=\pi$, and the tangent line equation is $y=\pi$ Step 4. See graph.