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=-1/2x=+-2Π/3+2Πn, n€ZPешим с помощью двойного неравенства:1) -Π/2<=-2Π/3+2Πn<=Π/2-Π/2+2Π/3<=2Πn<=Π/2+2Π/3Π/6<=2Πn<=7Π/6Π/12<=Πn<=7Π/121/12<=n<=7/12Нет 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<=1/12Нет целых чисел.Ответ: нет корней.