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

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

$y=\sin x(1+\cos x)=\sin x+\frac{1}{2} \sin 2 x$

$\Rightarrow \frac{d y}{d x}=\cos x+\cos 2 x=0$

∴  $\frac{d y}{d x}=0 \Rightarrow \cos 2 x=-\cos x=\cos (\pi-x)$

$\Rightarrow 2 x=\pi-x \Rightarrow x=\frac{\pi}{3}$

$\frac{d^2 y}{d x^2}=-\sin x-2 \sin 2 x<0$  for  $x=\frac{\pi}{3}$

∴  y is maximum at $x=\frac{\pi}{3}$ and its value is $\frac{3 \sqrt{3}}{4}$.