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 = π/6 + 2πn, n ∈ Z

х2 = 5π/6 + 2πn, n ∈ Z

2) sinx = - 1

x3 = - π/2 + 2πn, n ∈Z