What is the absolute minimum value of $f(x)=x-2\sin x$ on $[0,2\pi]$?

Core Concept

$f'(x)=1-2\cos x$ is zero when $\cos x=\frac{1}{2}$, at $x=\pi/3$ and $x=5\pi/3$. Candidates: $f(0)=0$, $f(\pi/3)=\pi/3-\sqrt{3}$, $f(5\pi/3)=5\pi/3+\sqrt{3}$, $f(2\pi)=2\pi$. The smallest is $\pi/3-\sqrt{3}$.