How to find the local maximum and minimum of $f(x)=x+\sin(2x)$ bounded by the interval $-\frac{2\pi}{3}$ and $\frac{2\pi}{3}$? 
Find the local maximum and minimum of $y=x+\sin(2x)$ between
$-\frac{2\pi}{3}$  and $\frac{2\pi}{3}$.

What I have done so far:
$f'(x) = 1+2\cos(2x)$  and $f''(x)=-4\sin(2x)$.

First, find where the derivative is zero ($y'=1+2\cos(2x)=0$).
$$1+2\cos(2x)=0$$
$$2\cos(2x)=-1$$
$$\cos(2x)=-\frac{1}{2}$$
$$2x=-\frac{2\pi}{3}, \frac{2\pi}{3}$$
(within the bounds)
$$x=-\frac{\pi}{3}, \frac{\pi}{3}$$
Checking the second derivative, you can find that $x=-\frac{\pi}{3}$ is local minimum and $x=\frac{\pi}{3}$ is local maximum.
