University Calculus: Early Transcendentals (3rd Edition)

The curve does have horizontal tangents at the $x-$axis. See below for explanations.
$$y=\sin(x-\sin x)$$ a) Find the derivative of the function: $$y'=\cos(x-\sin x)\times(x-\sin x)'$$ $$y'=\cos(x-\sin x)\times(1-\cos x)$$ $$y'=(1-\cos x)\cos(x-\sin x)$$ b) The curve would have horizontal tangents when there are values of $x$ for which its derivative function $y'=0$. Here as $y'=(1-\cos x)\cos(x-\sin x)$, we see that for $y'=0$, there is a case which $$1-\cos x=0$$ $$\cos x=1$$ $$x=k2\pi\hspace{1cm}(k\in Z)$$. This means the $x-$axis will act as the horizontal tangents to the curve at the points corresponding to $x=k2\pi$ with $k\in Z$.