If $0<x,y<\pi$ and $\cos x+\cos y-\cos \left(x+y\right)=\frac{3}{2}$, then $\sin x+\cos y$ is equal to:

$\cos x+\cos y-\cos \left(x+y\right)=\frac{3}{2}$

$2\cos \left(\frac{x+y}{2}\right)·\cos \left(\frac{x-y}{2}\right)-\left[2{\cos }^2\left(\frac{x+y}{2}\right)-1\right]=\frac{3}{2}$

$2\cos \left(\frac{x+y}{2}\right)·\cos \left(\frac{x-y}{2}\right)-2{\cos }^2\left(\frac{x+y}{2}\right)+1=\frac{3}{2}$

$2\cos \left(\frac{x+y}{2}\right)·\cos \left(\frac{x-y}{2}\right)-2{\cos }^2\left(\frac{x+y}{2}\right)=\frac{1}{2}$

$4\cos \left(\frac{x+y}{2}\right)·\cos \left(\frac{x-y}{2}\right)-4{\cos }^2\left(\frac{x+y}{2}\right)=1$

$\Rightarrow 4\cos \left(\frac{x+y}{2}\right)·\cos \left(\frac{x-y}{2}\right)-4{\cos }^2\left(\frac{x+y}{2}\right)=1={\sin }^2\left(\frac{x-y}{2}\right)+{\cos }^2\left(\frac{x-y}{2}\right)$

$\Rightarrow 4{\cos }^2\left(\frac{x+y}{2}\right)+{\cos }^2\left(\frac{x-y}{2}\right)-4\cos \left(\frac{x+y}{2}\right)·\cos \left(\frac{x-y}{2}\right)+{\sin }^2\left(\frac{x-y}{2}\right)=0$

$\Rightarrow {\left(2\cos \left(\frac{x+y}{2}\right)-\cos \left(\frac{x-y}{2}\right)\right)}^2+{\sin }^2\left(\frac{x-y}{2}\right)=0$

$\Rightarrow \sin \left(\frac{x-y}{2}\right)=0$ and $\cos \left(\frac{x+y}{2}\right)=\frac{1}{2}\cos \left(\frac{x-y}{2}\right)$

$\Rightarrow x=y$ and $\cos x=\frac{1}{2}=\cos y$

$\therefore \sin x=\frac{\sqrt{3}}{2}$

$\Rightarrow \sin x+\cos y=\frac{1+\sqrt{3}}{2}$

option $\left(4\right)$