Укажите корень уравнения cosx=-1/2 принадлежащий отрезку -p/2;p/2 а)p/4 b)корней нет c)p/2 d)p/6

cosx=-1/2 x=+-2Π/3+2Πn, n€Z Pешим с помощью двойного неравенства: 1) -Π/2<=-2Π/3+2Πn<=Π/2 -Π/2+2Π/3<=2Πn<=Π/2+2Π/3 Π/6<=2Πn<=7Π/6 Π/12<=Πn<=7Π/12 1/12<=n<=7/12 Нет целых чисел. 2) -Π/2<=2Π/3+2Πn<=Π/2 -Π/2-2Π/3<=2Πn<=Π/2-2Π/3 -7Π/6<=2Πn<=-Π/6 -7Π/12<=Πn<=Π/12 -7/12<=n<=1/12 Нет целых чисел. Ответ: нет корней.