решить уравнение sin х -sin2х=cos 2х

Вот: 2sin2х - sinx cosx - cos2x = 0 4sinx cosx - sinx cosx - cos2x = 0 3sinx cosx - ( cos^2x - sin^2x) = 0 sin^2x + 3 sinx cosx - cos^2x = 0 // : cos^2x ≠ 0 tg^2x + 3tgx - 1 = 0 D = 9 + 4 = 13 tgx = ( - 3 + √13)/2 ==> x = arctg (( - 3 + √13)/2 ) + pik, k ∈ Z tgx = ( - 3 - √13)/2 ==> x = arctg (( - 3 - √13)/2 ) + pik, k ∈ Z