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 = 0Пусть t = sinx, t ∈ [-1; 1].2t² - t - 1 = 0D = 1 + 4•2 = 3²t1 = (1 + 3)/4 = 1t2 = (1 - 3)/4 = -1/2Обратная замена:sinx = 1x = π/2 + 2πn, n ∈ Zsinx = -1/2x = (-1)ⁿ+¹π/6 + πn, n ∈ Z.2) sin2x + 2cosx = 02sinxcosx + 2cosx = 0cosx(sinx + 1) = 0cosx = 0x = π/2 + πn, n ∈ Zsinx = -1x = -π/2 + 2πn, n ∈ Z3) sin(2x - π/3) = -12x - π/3 = -π/2 + 2πn, n ∈ Z.2x = -π/6 + 2πn, n ∈ Zx = -π/12 + πn, n ∈ Z