Find all the points, if any, such that the curve $y=\sin(x-\sin x)$ has horizontal tangents at the $x$- axis.
My Solution
$\frac{\mathrm{d} y}{\mathrm{d} x}\sin(x-\sin x)=(\cos (x-\sin x))(1-\cos x)$
Let $\frac{\mathrm{d} y}{\mathrm{d} x}=0$ to find all the horizontal points.
We have, $(\cos (x-\sin x))(1-\cos x)=0$ which implies $(\cos (x-\sin x))=0$ or $(1-\cos x)=0$
Solving $(1-\cos x)=0$ first, we have $x= 2n\pi$ for all integers $n$.
Solving $(\cos (x-\sin x))=0$, I'm plain stuck. Any help thanks!