Помогите решить (1-√2 cosx)(1+sin4x)=0
(1-√2cosx)(1+sin4x)=0 a)1-√2cosx=0,√2cosx=1,cosx=1/√2,cosx=√2/2 x1=π/4+2kπ x2=7π/4+2kπ b)1+sin4x=0, sin4x=-1 4x=3π/2+2kπ, x3=3π/8+kπ/2 k∈Z Otvet: x1=π/4+2kπ,x2=7π/4+2kπ,x3=3π/8+kπ/2 , k∈Z