Cos2x+cos4x-cos3x=0 решите подробно

task/30401255  cos2x+cos4x - cos3x=0

решение   * * * cosα+cosβ=2cos( (α+β)/2 )*cos( (α-β)/2) * * *

cos4x+cos2x - cos3x= 0 ⇔ 2cos(3x) *cos(x)- cos(3x) = 0  ⇔

2cos(3x)* ( cos(x) -1/2 ) =0 ⇔ [ cos(3x)=0 ; cos(x) = 1/2 . ⇔

[3x=π/2 +πn , n∈ ℤ ; x  = ±  π/6 +2πк , k ∈ ℤ .

[ x = π/6 +(π/3)*n , n∈ ℤ ;  x  = ±  π/6 +2πк , k ∈ ℤ .

cos2x + cos4x - cos3x =0   2cos3x*cosx-cos3x=0 cos3x=0     x=П/6+П/3n   2cosx=1 cosx=1/2 x=плюс/минус П/3+2Пn