Решение
2cos(2x)=-√2
cos(2x)=-√2/2
cos(2π-2x)=-√2/2
2x=arccos(-√2/2)
2π-2x=arccos(-√2/2)
2x=3π/4
2π-2x=3π/4
2x=3π/4+2kπ,k∈Z
2π-2x=3π/4+2kπ,k∈Z
x=3π/8+kπ,k∈Z
x=5π/8-kπ,k∈Z
x=5π/8+kπ,k∈Z
Ответ
х={ 3π/8 +kπ ,k∈Z
{5π/8 +kπ