max _ 0 ≤ x ≤ π \ x - 2 x x + (1)/(3) 3x\ =

Let $f\left(x\right) = x - 2\sin x \cos x + \dfrac{1}{3}\sin 3x$

$⇒ {f}^{'}\left(x\right) = 1 - 2\cos 2x + \cos 3x$

$⇒ {f}^{"}\left(x\right) = 4\sin 2x - 3\sin 3x$

For maxima/minima ${f}^{'}\left(x\right) = 0$

$⇒ 1 - 2\left(2\cos^{2}x - 1\right) + 4\cos^{3}x - 3\cos x = 0$

$⇒ ( 2\cos x + \sqrt{3} ) ( 2\cos x - \sqrt{3} ) ( \cos x - 1 ) = 0$

$\cos x = \dfrac{- \sqrt{3}}{2}, \dfrac{\sqrt{3}}{2}, 1$

$x = \dfrac{5\pi}{6}, \dfrac{\pi}{6}, 0$

${f}^{"}\left(\dfrac{5\pi}{6}\right) = - 2\sqrt{3} - \sqrt{3} < 0$

${f}^{"}\left(\dfrac{\pi}{6}\right) = 2\sqrt{3} - \sqrt{3} > 0$

${f}^{"} ( 0 ) = 0$

So $x = \dfrac{5\pi}{6}$is local maxima point

Maximum value of $f\left(x\right) = f\left(\dfrac{5\pi}{6}\right) = \dfrac{5\pi}{6} + \dfrac{\sqrt{3}}{2} + \dfrac{1}{3}$

$= \dfrac{5\pi + 2 + 3\sqrt{3}}{6}$

Hence this is the correct option.