2sin^2-2cos2x-sin2x=0 , укажите корни на отрезке -6п -9п/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 <= n <= 1; n = 1 => x = pi + 2pi = 3pi; n = 0 => x = pi; pi <= - 2pi/3 + 2pi * n <= 3pi; pi + 2pi/3 <= 2pin <= 3pi + 2pi/3; 5pi/3 <= 2pin <= 11pi/3; 5/6 <= n <= 11/6, n = 1 => x = 4pi/3; pi <= 2pi/3 + 2pi * n <= 3pi, pi/3 <= 2pi * n <= 7pi/3; 1/6 <= n <= 7/6; n = 1 => x = 2pi/3 + 2pi = 8pi/3. ответ: pi, 3pi, 4pi/3, 8pi/3.