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