уравнения :)) досконально :)2sin x-1=0cos(2x+П/6)+1=06sin в квадрат x-5 cos x+5=0

Решение
1)  2sin x-1=0
sinx = 1/2
x = (-1)^n arcsin(1/2) + k, kZ
x = (-1)^n (/6) + k, kZ
2)  cos(2x+П/6)+1=0
cos(2x+П/6) = - 1
2x+П/6 =  + 2n, nZ
2x =  - /6 + 2n, nZ
2x = 5/6 + 2n, nZ
x = 5/12 + n, nZ
3)  6sinx - 5cosx + 5 = 0
6(1 - cosx) - 5cosx + 5 = 0
6 - 6cosx - 5cosx + 5 = 0
6cosx + 5cosx - 11 = 0
cosx = t, ItI 1
6t + 5t - 11 = 0
D = 25 + 4*6*11 = 289
t = (- 5 - 17)/12
t = - 22/12
t = -11/6
t = - 1 (5/6) не удовлетворяет условию ItI  1
t = (- 5 + 11)/12
t = 1/2
cosx = 1/2
x = (+ -)arccos(1/2) + 2m, mZ
 x = (+ -) *(/3) + 2m, mZ