Question

The solution set of the system of equations, $x+y=\frac{2\pi }{3},\cos x+\cos y=\frac{3}{2}$, where $x$ & $y$ are real, is ____

• A. $0$
• B. $\varphi$
• C. $-\frac{1}{2}$
• D. None

Answer 1

Answer: (B)

$\varphi$

Given, $\cos \left(x\right)+\cos \left(y\right)=\frac{3}{2}$
$\cos \left(x\right)+\cos (\frac{2\pi }{3}-x)=\frac{3}{2}$
$\cos \left(x\right)+\cos \left(\frac{2\pi }{3}\right)\cos \left(x\right)+\sin \left(\frac{2\pi }{3}\right).\sin \left(x\right)=\frac{3}{2}$.
$\cos \left(x\right)-\frac{\cos \left(x\right)}{2}+\frac{\sqrt{3}.\sin \left(x\right)}{2}=\frac{3}{2}$
$\frac{\cos \left(x\right)+\sqrt{3}\sin \left(x\right)}{2}=\frac{3}{2}$.
$\sin (x+\frac{\pi }{6})=\frac{3}{2}$
Now, $\sin (x+\frac{\pi }{6})$ cannot be greater than 1.
Hence, no solution.