Решите уравнение: cos2x=sinx

1 - 2sin^2x = sinx
- 2sin^2x - sinx + 1 = 0  /*(-1)
2sin^2x + sinx - 1 = 0 
sinx = t
2t^2 + t - 1 = 0 
D = 1 + 4*2 = 9 = 3^2
t1 = ( - 1 + 3)/4 = 2/4 = 1/2
t2 = ( - 1  - 3)/4 = - 4/4 = - 1

1) sinx = 1/2
х1 = /6 + 2n,    n  Z
х2 = 5/6 + 2n,    n  Z

2) sinx = - 1
x3 = - /2 + 2n, n Z