Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 141: 42

Yes, $x=\pi/6, 5\pi/6$

Step 1. Given $y=x+2cos(x)$, we have $y'=1-2sin(x)$ Step 2. For a horizontal tangent, let $y'=0$, we have $sin(x)=1/2$ and $x=\pi/6, 5\pi/6$ in $[0,2\pi]$ Step 3. At $x=\pi/6$, $y=\pi/6+2cos(\pi/6)=\pi/6+\sqrt 3$, and the tangent line equation is $y=\pi/6+\sqrt 3$ Step 4. At $x=5\pi/6$, $y=5\pi/6+2cos(5\pi/6)=\pi/6-\sqrt 3$, and the tangent line equation is $y=\pi/6-\sqrt 3$ Step 5. See graph.