For the smallest positive values of x and y, the equation $2(\sin \mathrm{x}+\sin \mathrm{y})-2\cos (\mathrm{x}-\mathrm{y})=3$has a solution, then which of the following is/are true?

The given equation is $2(\sin \mathrm{x}+\sin \mathrm{y})-2\cos (\mathrm{x}-\mathrm{y})=3$
$\Rightarrow 2\times 2\sin \frac{\mathrm{x}+\mathrm{y}}{2}\cos \frac{\mathrm{x}-\mathrm{y}}{2}-2\left[2{\cos }^2 \frac{\mathrm{x}-\mathrm{y}}{2}-1\right]=3$

or   $4{\cos }^2 \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)-4\sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)+1=0$
or $\cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\frac{4\sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\pm \sqrt{16{\sin }^2 \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)}-16}{8}$
$\therefore {\sin }^2\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\ge 1\Rightarrow \sin \frac{\mathrm{x}+\mathrm{y}}{2}=\pm 1$
Since x and y are smallest and positive, we have
$\sin \frac{\mathrm{x}+\mathrm{y}}{2}=1\text{ and }\frac{\mathrm{x}+\mathrm{y}}{2}=\frac{\mathrm{\pi }}{2}$
i.e., $\mathrm{x}+\mathrm{y}=\mathrm{\pi } \left(\mathrm{i}\right)$
Also,  $\cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)=\frac{1}{2}$
$\Rightarrow \mathrm{x}-\mathrm{y}=\frac{2\mathrm{\pi }}{3}\text{ or }-\frac{2\mathrm{\pi }}{3} \left(\mathrm{ii}\right)$
From Eqs. (i) and (ii), we get
$(\mathrm{x}=5\mathrm{\pi }/6,\mathrm{y}=\mathrm{\pi }/6)\text{ or }(\mathrm{x}=\mathrm{\pi }/6,\mathrm{y}=5\mathrm{\pi }/6)$