Find the number of pairs (x, y) satisfying the equation 

$\sin x+\sin y=\sin \left(x+y\right)and\left|x\right|+\left|y\right|=1$

 first equation can be written as$2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x+y}{2}\right)$

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

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

$\frac{x+y}{2}=n\pi or \frac{x}{2}=n\pi or\frac{y}{2}=n\pi$  $\Rightarrow y=2n\pi and given \left|x\right|+\left|y\right|=1$

$$\begin{gathered} \therefore \left|x\right|\le 1,\left|y\right|\le 1 \\ Hence, x + y = 0 Or x = 0 or y = 0 clearly y = 0 cuts the curve \\ \left|x\right|+\left|y\right|=1at A, C, x = 0, cuts the curve\left|x\right|+\left|y\right|=1 \\ at B, D and x + y = 0 cuts the curve at E, F, hence 6 solutions are possible \end{gathered}$$