A) 2
B) 4
C) 6
D) \[\infty \]
Correct Answer: C
Solution :
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 \[(0,\text{ }\pm 1)\] and solving with\[y=0\], we get \[(\pm \text{ }1,\,\,0)\] as the other solution. Thus we have six pairs of solution for \[-4,\,-3,\,-2,\,-1,\,0,\,1,\,2,\,3\]and\[y\].
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