2sin^2-2cos2x-sin2x=0, укажите корни на отрезке - 6 п - 9 п/2

2sin^2x - 3cosx - 3 = 0, х ∈ [pi; 3pi]; 2 (1 - cos^2x) - 3cosx - 3 = 0; 2 - 2cos^2x - 3cosx - 3 = 0; - 2cos^2x - 3cosx - 1 = 0; 2cos^2x + 3cosx + 1 = 0; Пусть cosx = t, тогда 2t^2 + 3t + 1 = 0; D = 9 - 4 * 2 * 1 = 1; t = (-3 + - 1) / (2 * 2) ; t1 = - 1/2, t2 = - 1; cosx = - 1/2, x = + - arccos (-1/2) + 2pi * n, n ∈ N, x = + - 2pi/3 + 2pi * n, n ∈ N; cosx = - 1, x = pi + 2pin, n ∈ N; pi < = pi + 2pin < = 3pi; 0 < = 2pin < = 2pi; 0 <+> x = pi; pi < = - 2pi/3 + 2pi * n < = 3pi; pi + 2pi/3 < = 2pin < = 3pi + 2pi/3; 5pi/3 < = 2pin < = 11pi/3; 5/6 < = n x = 4pi/3; pi < = 2pi/3 + 2pi * n < = 3pi, pi/3 < = 2pi * n < = 7pi/3; 1/6 < = n x = 2pi/3 + 2pi = 8pi/3. Ответ: pi, 3pi, 4pi/3, 8pi/3.