Решите ! 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