QPaperGen

Let
\[
f(x) = x + \sin 2x
\]

Differentiate:
\[
f'(x) = 1 + 2\cos 2x
\]

Set derivative to zero for critical points:
\[
f'(x) = 0 \Rightarrow 1 + 2\cos 2x = 0 \Rightarrow \cos 2x = -\frac{1}{2}
\]

Now solve \( \cos 2x = -\frac{1}{2} \):
\[
2x = \frac{2\pi}{3}, \frac{4\pi}{3} \Rightarrow x = \frac{\pi}{3}, \frac{2\pi}{3}
\]

Also in the interval \([0, 2\pi]\), the values of \(x\) satisfying \( \cos 2x = -\frac{1}{2} \) are:
\[
x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}
\]

Evaluate \(f(x)\) at critical points and endpoints:
\[
f\left(0\right) = 0 + \sin 0 = 0
\]
\[
f\left(2\pi\right) = 2\pi + \sin 4\pi = 2\pi + 0 = 2\pi
\]
\[
f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \sin\left(\frac{2\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}
\]
\[
f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} + \sin\left(\frac{4\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}
\]
\[
f\left(\frac{4\pi}{3}\right) = \frac{4\pi}{3} + \sin\left(\frac{8\pi}{3}\right) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}
\]
\[
f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sin\left(\frac{10\pi}{3}\right) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2}
\]

Compare all values. The smallest is at \(x = 0\), and the largest is at \(x = 2\pi\).

\[
\boxed{
\text{Maximum value of } f(x) \text{ on } [0, 2\pi] \text{ is } 2\pi \text{ at } x = 2\pi
}
\]
\[
\boxed{
\text{Minimum value of } f(x) \text{ on } [0, 2\pi] \text{ is } 0 \text{ at } x = 0
}
\]