Ответы на вопрос:
Sin²x-cos²x=cos(x/2) -(cos²x-sin²x)=cos(x/2) -cos2x-cos(x/2)=0 cosα+cosβ=2cos((α+β)/2) *cos((α-β)/2) -(cos2x+cos(x/2))=0 2cos(5/4)x *cos(3/4)x=0 cos(5/4)x=0 или cos(3/4)x=0 1. cos(5/4)x=0 5/4x=π/2+πn, n∈z x₁=2π/5+4πn/5, n∈z 2. cos(3/4)x=0 3/4x=π/2+πn, n∈z x₂=2π/3+4πn/3, n∈z
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