Sin^2x+sin2x=1 не могу решить ((

sin^2x + 2sinxcosx - (cos^2x + sin^2x) = 0  sin^2x + 2sinxcosx - cos^2x - sin^2x = 0  cosx (2sinx - cosx) = 0  1) cosx =  0 x = pi/2 + pik, k ∈Z 2) 2sinx - cosx = 0   /:cosx≠ 0 2tgx - 1 = 0  tgx = 1/2 x = arctg(1/2) + pik, k ∈Z

sin²x + sin2x = 1 Разложим синус удвоенного аргумента: sin²x + 2sinxcosx - 1 = 0 -(1 - sin²x) + 2sinxcosx = 0 -cos²x + 2sinxcosx = 0 cosx(-cosx + 2sinx) = 0 cosx = 0 x = π/2 + πn, n ∈ Z cosx = 2sinx 1 = 2tgx tgx = 1/2 x = arctg(1/2) + πn, n ∈ Z.