решите уравнение sin2x-2√3cos(x+7п/6)=3cosx

sin2x - 2√3•cos(x + 7π/6) = 3cosx

cos(α + β) = cosα•cosβ - sinα•sinβ

cosx•cos(7π/6) - sinx•sin(7π/6) = - cosx•cos(π/6) + sinx•sin(π/6) = - (√3/2)•cosx + (1/2)•sinx

sin2x - 2√3•( - (√3/2)•cosx + (1/2)•sinx ) = 3cosxsin2x + 3cosx - √3sinx = 3cosx

sin2x = 2sinx•cosx

2sinx•cosx - √3sinx = 0sinx•(2cosx - √3) = 01)  sinx = 0  ⇔  x = πn , n ∈ Z2)  2cosx - √3 = 0  ⇔  cosx = √3/2  ⇔  x = ± π/6 + 2πk , k ∈ ZОТВЕТ: πn , n ∈ Z ; ± π/6 + 2πk , k ∈ Z