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 .
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 .