Trigonometry related doubt

Question

the number of all possible ordered pairs (x,y),x,y belongs to R satisfying the system of equations x+y=2pi/3,cosx +cosy=3/2

Answer 1

y = 2π/3 - x

 

So, cosx + cosy = cosx + cos(2π/3 - x) = cosx + cos(2π/3)cosx + sin(2π/3)sinx = 3/2

 

cosx - (1/2)cosx + (√3/2)sinx = 3/2

 

(1/2)cosx + (√3/2)sinx = 3/2

 

cos(π/3)cosx + sin(π/3)sinx = 3/2

 

cos(π/3 - x) = 3/2

 

cos(π/3 - x) can't be larger than 1, so there are no solutions.