1. Let $f(x)=x-2\sin x\cos x+\frac13\sin 3x, \qquad 0\le x\le \pi.$ We need to find $\max_{0\le x\le \pi} f(x).$

2. Simplify the trigonometric part.

Since $2\sin x\cos x=\sin 2x,$ we get $f(x)=x-\sin 2x+\frac13\sin 3x.$

3. Differentiate to locate critical points.

$f'(x)=1-2\cos 2x+\cos 3x.$

Now simplify this expression. Using $\cos 3x=4\cos^3x-3\cos x, \qquad \cos 2x=2\cos^2x-1,$

$f'(x)=1-2(2\cos^2x-1)+(4\cos^3x-3\cos x).$

So, $f'(x)=1-4\cos^2x+2+4\cos^3x-3\cos x$ $=4\cos^3x-4\cos^2x-3\cos x+3.$

Factor this: $4c^3-4c^2-3c+3=(c-1)(4c^2-3), \quad c=\cos x.$

Hence $f'(x)=(\cos x-1)(4\cos^2x-3).$

4. Solve $f'(x)=0$ in $[0,\pi]$.

Either $\cos x=1 \Rightarrow x=0,$ or $4\cos^2x-3=0 \Rightarrow \cos^2x=\frac34 \Rightarrow \cos x=\pm \frac{\sqrt3}{2}.$

Thus the critical points in $[0,\pi]$ are $x=0,\ \frac\pi6,\ \frac{5\pi}6.$

Also check the endpoint $x=\pi$.

5. Evaluate $f(x)$ at all relevant points.

• At $x=0$: $f(0)=0-0+0=0.$

• At $x=\frac\pi6$: $f\left(\frac\pi6\right)=\frac\pi6-\sin\frac\pi3+\frac13\sin\frac\pi2$ $=\frac\pi6-\frac{\sqrt3}{2}+\frac13.$

• At $x=\frac{5\pi}6$: $f\left(\frac{5\pi}6\right)=\frac{5\pi}6-\sin\frac{5\pi}3+\frac13\sin\frac{5\pi}2.$

Now, $\sin\frac{5\pi}3=-\frac{\sqrt3}{2}, \qquad \sin\frac{5\pi}2=1.$ Therefore, $f\left(\frac{5\pi}6\right)=\frac{5\pi}6-\left(-\frac{\sqrt3}{2}\right)+\frac13$ $=\frac{5\pi}6+\frac{\sqrt3}{2}+\frac13.$

Putting over denominator $6$, $f\left(\frac{5\pi}6\right)=\frac{5\pi+3\sqrt3+2}{6}.$

• At $x=\pi$: $f(\pi)=\pi-0+0=\pi.$

6. Compare the values.

We have $f(0)=0,$ $f\left(\frac\pi6\right)=\frac\pi6-\frac{\sqrt3}{2}+\frac13,$ $f\left(\frac{5\pi}6\right)=\frac{5\pi+2+3\sqrt3}{6},$ $f(\pi)=\pi.$

Clearly, $\frac{5\pi+2+3\sqrt3}{6}>\pi$ (since this is equivalent to $2+3\sqrt3>\pi$, which is true). So the maximum occurs at $x=\frac{5\pi}6.$

7. Therefore, $\max_{0\le x\le \pi}\left\{x-2\sin x\cos x+\frac13\sin 3x\right\} = \frac{5\pi+2+3\sqrt3}{6}.$

Hence the correct option is A.