Уравнения :)) подробно :) 2sin x-1=0 cos(2x+П/6)+1=0 6sin в квадрат x-5 cos x+5=0

Решение 1)  2sin x-1=0 sinx = 1/2 x = (-1)^n arcsin(1/2) + πk, k∈Z x = (-1)^n (π/6) + πk, k∈Z 2)  cos(2x+П/6)+1=0 cos(2x+П/6) = - 1 2x+П/6 = π + 2πn, n∈Z 2x = π - π/6 + 2πn, n∈Z 2x = 5π/6 + 2πn, n∈Z x = 5π/12 + πn, n∈Z 3)  6sin²x - 5cosx + 5 = 0 6(1 - cos²x) - 5cosx + 5 = 0 6 - 6cos²x - 5cosx + 5 = 0 6cos²x + 5cosx - 11 = 0 cosx = t, ItI ≤ 1 6t² + 5t - 11 = 0 D = 25 + 4*6*11 = 289 t₁ = (- 5 - 17)/12 t₁ = - 22/12 t₁ = -11/6 t₁ = - 1 (5/6) не удовлетворяет условию ItI ≤ 1 t₂ = (- 5 + 11)/12 t₂ = 1/2 cosx = 1/2 x = (+ -)arccos(1/2) + 2πm, m∈Z  x = (+ -) *(π/3) + 2πm, m∈Z