$\cos (x-y)-2\sin x+2\sin y=3$ if

$\cos (x-y)-2\sin x+2\sin y=3$

$\Rightarrow 1-2{\sin }^2\frac{x-y}{2}-4\sin \frac{x-y}{2}\cos \frac{x+y}{2}-3=0$

$\begin{array}{l}\Rightarrow {\sin }^2\frac{x-y}{2}+2\sin \frac{x-y}{2}\cos \frac{x+y}{2}+1=0\\ \Rightarrow \sin \frac{x-y}{2}=\frac{-2\cos \frac{x+y}{2}}{2}\sqrt{4{\cos }^2\frac{x+y}{2}-4}\end{array}$

For real values of $\sin \frac{x-y}{2}$, we have ${\cos }^2\frac{x+y}{2}=1$

$\Rightarrow {\sin }^2\frac{x+y}{2}=0$ or $\sin \frac{x+y}{2}=0$

$\Rightarrow x+y=2n\pi$

and then $\sin \frac{x-y}{2}=1\Rightarrow \frac{x-y}{2}=(2k-1)\frac{\pi }{2}$

$\Rightarrow x-y=(2k-1)\pi$

so (b) is not correct.

Also if $\sin x=\sin y$ then the given equation becomes $\cos (x-y)=3$ which is not  correct;  (a) is not correct next if

$x=2k\pi -\pi /2$ and $y=2n\pi +\pi /2$.

Then $\cos (x-y)-2\sin x+2\sin y=-1+2+2=3,$so (c) is correct.

Finally  if $\cos (x-y)=-1$, the given  equation becomes  $\sin x-\sin y=-2$ which is not true for any real values  of x and y so (d) is not correct.