From the given equation we have,
$ 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}=2 \sin \frac{x+y}{2} \cos \frac{x+y}{2}$
$ \Rightarrow \sin \frac{x+y}{2}\left[\cos \frac{x-y}{2}-\cos \frac{x+y}{2}\right]=0 $
$ \Rightarrow \sin \frac{x+y}{2} \times 2 \sin \frac{x}{2} \sin \frac{y}{2}=0$
Which holds if either $\sin \frac{x+y}{2}=0$ or $\sin \frac{x}{2}=0$
Also since $|x|+|y|=1 \Rightarrow|x| \leq 1,|y| \leq 1$. So, the required solution is either $x+y=0$ or $x=0$ or $y=0$
If $x=0, y= \pm 1$, and if $y=0, x= \pm 1$ and if $x+y=0$
$\Rightarrow \mathrm{y}=-\mathrm{x} \Rightarrow|\mathrm{y}|=|-\mathrm{x}|=|\mathrm{x}|=\frac{1}{2}$
So that the required pairs $(\mathrm{x}, \mathrm{y})$ are $(0, \pm 1),( \pm 1,0),\left(\frac{1}{2},-\frac{1}{2}\right),\left(-\frac{1}{2}, \frac{1}{2}\right)$ which are 6 in number.