Question

№6 помогите пожалуйста

Answer 1

из первого уравнения: х = (π/3) - у
подставим во второе: cos((π/3) - у) - 2cos(y) = 0
cos(π/3)*cos(y) + sin(π/3)*sin(y) - 2cos(y) = 0
(1/2)*cos(y) + (√3/2)*sin(y) - 2cos(y) = 0
cos(y) + √3*sin(y) - 4cos(y) = 0
√3*sin(y) - 3cos(y) = 0
sin(y) = √3cos(y)
tg(y) = √3
y = (π/3) + πk, k∈Z
х = (π/3) - (π/3) - πk, k∈Z ---> x = πk, k∈Z