Find the maximum value of the function $sinx(1+cosx)$ is :

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

The correct answer is Option (1) → $\frac{3\sqrt{3}}{4}$

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

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

$=\cos x+\cos^2x-\sin^2x=0$

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

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

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

so $y_{max}=\frac{\sqrt{3}}{2}(1+\frac{1}{2})=\frac{3\sqrt{3}}{4}$