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 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.