Symmetry of g(x) = x - 2sin(x)

Substitute $-x$ for $x$: $g(-x) = (-x) - 2\sin(-x) = -x - 2(-\sin x) = -x + 2\sin x$. This can be rewritten as $g(-x) = -(x - 2\sin x) = -g(x)$.

Since $g(-x) = -g(x)$ for all $x$ in the domain, the function satisfies the definition of an odd function. Therefore $g(x) = x - 2\sin(x)$ is odd, meaning its graph is symmetric about the origin.

Differentiate: $g'(x) = 1 - 2\cos(x)$. Setting $g'(x) = 0$ gives $\cos(x) = \tfrac{1}{2}$, so on $[-2\pi, 2\pi]$ the solutions are $x = \pm\tfrac{\pi}{3}, \pm\tfrac{5\pi}{3}$.

Testing the sign of $g'(x) = 1 - 2\cos(x)$ on intervals between these points shows $g'$ changes from negative to positive at $x = -\tfrac{5\pi}{3}$ and $x = \tfrac{\pi}{3}$ (local minima), and from positive to negative at $x = -\tfrac{\pi}{3}$ and $x = \tfrac{5\pi}{3}$ (local maxima).