Find the number of pairs (x, y) satisfying the equation  $\sin x+\sin y=\sin (x+y)and\left|x\right|+\left|y\right|=1$

The first equation can be written as 

$$\begin{gathered} 2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x+y}{2}\right) \\ \Rightarrow 2\sin \left(\frac{x+y}{2}\right)\{\cos \left(\frac{x-y}{2}\right)-\cos \left(\frac{x+y}{2}\right)\}=0 \\ \Rightarrow 2\sin \left(\frac{x+y}{2}\right)2\sin \left(\frac{x}{2}\right)\sin \left(\frac{y}{2}\right)=0 \\ Either \frac{x+y}{2}=n\pi or \frac{x}{2}\text{=}n\pi or\frac{y}{2}=n\pi \\ Either x+y=2n\pi orx=2n\pi ory=2n\pi \\ \therefore \left|x\right|+\left|y\right|=1 \end{gathered}$$

$$\begin{gathered} \therefore \left|x\right|\le 1 \\ and\left|y\right|\le 1 \end{gathered}$$

Hence,    $x + y = 0$
or        $x = 0 or y = 0$ clearly $y = 0$ cuts the curve  $\left|x\right|+\left|y\right|=1$ at A, C, $x = 0$, cuts the curve  $\left|x\right|+\left|y\right|=1$ at B, D and $x + y = 0$  cuts the curve at E, F, hence 6 solutions are possible.