Ребят, просьба помочь решить 10 тригонометрических уравнений.Ошибаться нельзя-решается судьба годовой оценки.

1)Cos(4x - π/3) = 1 4x - π/3 = 2πk, k ∈Z 4x = 2πk + π/3, k ∈Z x = πk/2 + π/12, k∈Z 2)Cos(π/6 +3х) = 0 π/6 +3х = π/2 + πk , к ∈Z 3x = π/2 + πk - π/6, k ∈Z 3x = π/3 + πk, k ∈Z x = π/9 + πk/3, k ∈Z 3)Sin(x/2 - π/3) = 1/2 х/2 - π/3 = (-1)^n arcSin1/2 + nπ, n ∈Z' x/2 -π/3 = (-1)^n π/6 + πn, n ∈Z x/2 = (-1)^n π/6 + πn + π/3, n ∈Z x = (-1)^n π/3 + 2πn + 2π/3 , n ∈Z 4)Sin(2x - π/6) = 0 2х - π/6 = nπ, n ∈Z 2x = nπ + π/6, n∈Z x = nπ/2 + π/12, n ∈Z 5)Sin(2x -π/3) = -1 2х - π/3 = -π/2 + 2πk ,к ∈Z 2x = π/3  -π/2 + 2πk, k ∈Z x = π/6  -π/4 + πk, k ∈Z 6) 2Sin² x +3Cosx = 0 2(1-Cos²x) + 3Cosx = 0 2 - 2Cos²x + 3Cosx = 0 2Cos²x -3Cosx - 2 = 0 Решаем как квадратное а)Соs x = 2 (нет решений) б) Сos x = -1/2 х = +-arcCos(-1/2) + 2πk, k ∈Z x = +- 2π/3 + 2πk, k ∈Z 7)Cos(4x + π/3) = 1 4x + π/3 = 2πk, k ∈Z 4x = 2πk - π/3, k ∈Z x = πk/2 - π/12, k∈Z 8) Sin(2x + π/4) = 1 2х + π/4 = π/2 + 2πк, к ∈Z 2x = -π/4 +π/2 + 2πк, к ∈Z x = -π/8 + π/4+ πк, к ∈Z 9) Cos(3x - π/6) = -1 3x - π/6 = π + 2πк , к ∈Z 3x = π/6 + π + 2πк , к ∈Z 3x =7π/6 + 2πк , к ∈Z x = 7π/18 +2πк/3, к ∈Z 10) Sin(x/2 + 11π/28) = 0 х/2  + 11π/28 = nπ, n ∈Z x/2 = nπ - 11π/28, n∈Z x =2 nπ - 11π/14, n ∈Z