A) \[\left( \frac{2}{3},\frac{1}{3} \right)\]
B) \[\left( \frac{5}{3},\frac{4}{3} \right)\]
C) \[\left( \frac{13}{6},\frac{11}{6} \right)\]
D) \[\left( \frac{5}{12},\frac{1}{12} \right)\]
Correct Answer: C
Solution :
\[{{\cos }^{2}}\pi x-{{\sin }^{2}}\pi y=\frac{1}{2}\] \[\cos \pi (x+y)\cdot cos\left( \frac{\pi }{3} \right)=\frac{1}{2}\] \[\cos (\pi )\cdot (x+y)=1\] \[\Rightarrow \pi (x+y)=2n\pi ,\]where\[n\in I\] \[\Rightarrow x+y=2n\]and \[x-y=\frac{1}{3}\] \[\Rightarrow x=n+\frac{1}{6};y=n-\frac{1}{6};(n\in I)\] \[(x,y)\in \left( n+\frac{1}{6},n-\frac{1}{6} \right)\]which is satisfied by 3rd option for\[n=2\].
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