решите уравнение sin2x=|cos x|

sin2x = | cosx |Первый случай, когда cosx ≥ 0, тогдаsin2x = cosx2sinx × cosx - cosx = 0cosx × ( 2sinx - 1 ) = 0________________cosx = 0x = π/2 + πn, n € Z________________2sinx - 1 = 0sinx = 1/2x = π/6 + 2πk, k € Zx = 5π/ 6 + 2πm, m € Z_________________Второй случай, когда соsx ≤ 0, тогдаsin2x = - cosx2sinx × cosx + cosx = 0cosx × ( 2sinx + 1 ) = 0________________cosx = 0х = π/2 + πn, n € Z________________2sinx + 1 = 0sinx = -1/2x = - π/6 + 2πl, l € Zx = -5π/ 6 + 2πs, s € Z_________________Ответ: 1) Если cosx ≥ 0, тох = π/2 + πn, n € Zx = π/6 + 2πk, k € Zx = 5π/ 6 + 2πm, m € Z2) Если cosx ≤ 0 , тох = π/2 + πn, n € Zx = - π/6 + 2πl, l € Zx = -5π/ 6 + 2πs, s € Z