Question

The number of ordered solution pairs $(x,y)$ in $[0,2\pi ]$ for the system of the equations $x+y=\frac{2\pi }{3}$ and $\cos x+\cos y=\frac{3}{2}$ is

Answer 1

$\cos x+\cos y=\frac{3}{2}$
$\Rightarrow 2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{3}{2}$
$\Rightarrow 2\cos \left(\frac{\pi }{3}\right)\cos \left(\frac{x-y}{2}\right)=\frac{3}{2}$
$\Rightarrow \cos \left(\frac{x-y}{2}\right)=\frac{3}{2}$
As $\cos \theta \in [-1,1]$, so there is no solution of the given equation.