Найдите корни уравнения 2cos x-1=0, принадлежащие отрезку [0;2п ]

2cosx - 1 = 0   [0; 2pi]
2cosx = 1 
cosx = 1/2

x = +- pi/3 + 2pik, k ∈ Z

1) k = -1
x1 = + pi/3 - 2pi = pi/3 - 6pi/3 = (pi-6pi)/3 = -5pi/3 ∉
x2 = - pi/3 - 2pi = - pi/3 - 6pi/3 = (-pi-6pi)/3 = -7pi/3 ∉
2) k = 0
x1 = +pi/3 ∈
x2 = - pi/3 ∉
3) k =1
x1 = +pi/3 + 2pi = pi/3 + 6pi/3 = (pi+6pi)/3 = 7pi/3 ∉
x2 = -pi/3 + 2pi = -pi/3 + 6pi/3 = (-pi+6pi)/3 = 5pi/3 ∈
4) k = 2
x1 = +pi/3 + 4pi = pi/3 + 12pi/3 = (pi+12pi)/3 = 13pi/3 ∉
x2 = -pi/3 + 4pi = -pi/3 + 12pi/3 = (-pi+12pi)/3 = 11pi/3 ∉

ОТВЕТ: pi/3; 5pi/3