Question

cosx+cos2x+cos3x+cos4x=0

Answer 1

cosα+cosβ=2(cos(α+β)/2)*cos((α-β)/2)
cosx+cos3x=2cos((x+3x)/2)*cos((x-3x)/2)=2cos2x*cosx
(функция у=cosx -четная функция)
cos2x+cos4x=2cos((2x+4x)/2)*cos(2x-4x)/2)=2cos3x*cosx
2cos2x*cosx+2cos3x*cosx=0
2cosx(cos2x+cos3x)=0
2cosx*2cos2,5x*cos0,5x=0
cosx=0, x₁=π/2+πn, n∈Z
cos2,5x=0, 2,5x=π/2+πn, ∈Z.  x₂=π/5+2π/5, n∈Z
cos0,5x=0, x=π/2+πn, n∈Z. x₃=π+2πn, n∈Z