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Решение

2cos(2x)=-√2

cos(2x)=-√2/2

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=3π/8+kπ,k∈Z

x=5π/8+kπ,k∈Z

ответ

х={ 3π/8 +kπ               ,k∈Z

   {5π/8 +kπ