(c) The first equation can be written as

$2\sin \frac{1}{2}(x + y)\cos \frac{1}{2}(x – y) = 2\sin \frac{1}{2}(x + y)\cos \frac{1}{2}(x + y)$

$\therefore $ Either $\sin \frac{1}{2}(x + y) = 0$ or $\sin \frac{1}{2}x = 0$ or $\sin \frac{1}{2}y = 0$

Thus $x + y = – 1,\,\,x – y = – 1$.

When $x + y = 0,$ we have to reject $x + y = 1$ and check with the options or $x + y = – 1$ and solve it with $x – y = 1$ or $x – y = – 1$

which gives $\left( {\frac{1}{2},\,\, – \frac{1}{2}} \right)$ or $\left( { – \frac{1}{2},\,\frac{1}{2}} \right)$ as the possible solution.

Again solving with $x = 0$, we get $(0,{\rm{ }} \pm 1)$ and solving with $y = 0$, we get $( \pm {\rm{ }}1,\,\,0)$ as the other solution.

Thus we have six pairs of solution for $ x$ and $y$.