Ответы на вопрос:
1cos3x=-2cosx 4cos³x-3cosx+2cosx=0 4cos³x-cosx=0 cosx(4cos²x-1)=0 cosx=0⇒x=π/2+πn,n∈z 4cos²x-1=0 4(1+cos2x)/2=1 1+cos2x=1/2 cos2x=-1/2 2x=+-2π/3+2πk x=+-π/3+πk,k∈z 2 cosx-cos(π/2-11x)=0 -2sin(6x-π/4)sin(-5x+π/4)=0 2sin(6x-π/4)sin(5x-π/4)=0 sin(6x-π/4)=0 6x-π/4=πn 6x=π/4+πn x=π/24+πn/6,n∈z sin(5x-π/4)=0 5x-π/4=πk 5x=π/4+πk x=π/20+πk/5,k∈z 3 (sinx+sin7x)+(cos3x-cos5x)=0 2sin4xcos3x+2sin4xsinx=0 2sin4x*(cos3x+sinx)=0 sin4x=0 4x=πn x=πn/4 cos3x+cos(π/2-x)=0 2cos(x+π/4)cos(2x-π/4)=0 cos(x+π/4)=0 x+π/4=π/2+πk x=π/4+πk,k∈z cos(2x-π/4)=0 2x-π/4=π/2+πm 2x=3π/4+πm x=3π/8+πm/2,m∈z
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