Решите уравнение! Cрочно нужно!

1) 2sin^2 (x) - sin (x) - 1 = 0

2) sin (2x) + 2cos (x) = 0

3) sin (2x - Pi/3) = - 1

1) 2sin²x - sinx - 1 = 0

Пусть t = sinx, t ∈ [-1; 1].

2t² - t - 1 = 0

D = 1 + 4•2 = 3²

t1 = (1 + 3) / 4 = 1

t2 = (1 - 3) / 4 = - 1/2

Обратная замена:

sinx = 1

x = π/2 + 2πn, n ∈ Z

sinx = - 1/2

x = (-1) ⁿ+¹π/6 + πn, n ∈ Z.

2) sin2x + 2cosx = 0

2sinxcosx + 2cosx = 0

cosx (sinx + 1) = 0

cosx = 0

x = π/2 + πn, n ∈ Z

sinx = - 1

x = - π/2 + 2πn, n ∈ Z

3) sin (2x - π/3) = - 1

2x - π/3 = - π/2 + 2πn, n ∈ Z.

2x = - π/6 + 2πn, n ∈ Z

x = - π/12 + πn, n ∈ Z