sin3x-cosx=cos3x-sin5x, cos3x-sin5x≠0
sin3x+sin5x=cos3x+cosx
2sin4xcosx=2cos2xcosx,
2cosx(sin4x-cos2x)=0
cosx=0, 2sin2xcos2x-cos2x=0, cos2x(2sin2x-1)=0
x1=π/2+πn,
cos2x=0, sin2x=0.5
2x=π/2+πn 2x=((-1)^n)π/6+πn
x2=π/4+πn/2 x3=((-1)^n)π/12+πn/2
Ответ: x1=π/2+πn, x2=π/4+πn/2, x3=((-1)^n)π/12+πn/2, n∈Z