The maximum value of the function $\sin x(1+\cos x)$ is

$y=\sin x(1+\cos x)$

$=\sin x+\frac{1}{2}\sin 2x$

$y\text{'}=\cos x+\cos 2x=0$

$2{\cos }^{2}x+\cos x-1=0$

$2{\cos }^{2}x+2\cos x-\cos x-1=0$

$2\cos x(\cos x+1)-1(\cos x+1)=0$

$\cos x=\frac{1}{2},\cos x=-1$

$\frac{d^{2}y}{d{x}^2}=\sin x-2\sin 2x$

$\left(\frac{{\displaystyle d^{2}y}}{{\displaystyle d{x}^2}}\right)=\sin \frac{\pi }{3}-2\sin \frac{2\pi }{3}<0$

$x=\frac{\pi }{3}$

$\therefore$y’ is maximum when $x=\frac{\pi }{3}$

And its value is $\frac{\sqrt{3}}{2}\left(1+\frac{1}{2}\right)$

$=\frac{3\sqrt{3}}{4}$