A) \[x=\frac{\pi }{3}-n\pi ,y=n\pi \]
B) \[\phi \]
C) \[x=n\pi ,y=\frac{\pi }{3}-n\pi \]
D) None of these
Correct Answer: B
Solution :
We have \[\cos x+\cos y=\frac{3}{2}\] \[\Rightarrow \,\,2\cos \left( \frac{x+y}{2} \right)\cos \left( \frac{x-y}{2} \right)=\frac{3}{2}\] \[\Rightarrow \,\,\cos \left( \frac{x-y}{2} \right)=\frac{3}{2}\left( \because \,\,x+y=\frac{2\pi }{3} \right)\] Which is not possible (as \[\cos \theta \le 1\]) Thus, the solution set is a null set.
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