$f\left(x\right)=\sin x(1+\cos x)$  has maximum value at $x=$

 $f\left(x\right)=\sin x(1+\cos x)$

$f\text{'}\left(x\right)=\cos x\left(1+\cos x\right)+\sin x\left(-\sin x\right)=\cos x+\cos 2x$

$f\text{'}\text{'}\left(x\right)=-2\sin 2x-\sin x$

$f\text{'}\left(x\right)=0\Rightarrow \cos 2x+\cos x=0$

$2\cos \frac{3x}{2}\cos \frac{x}{2}=0$

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

$f\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=0, f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{3}\right)<0$

f has max value at $x=\frac{\mathrm{\pi }}{3}$