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

2cosx - 1 = 0   [0; 2pi]2cosx = 1 cosx = 1/2x = +- pi/3 + 2pik, k ∈ Z1) k = -1x1 = + 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 = 0x1 = +pi/3 ∈x2 = - pi/3 ∉3) k =1x1 = +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 = 2x1 = +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