Решите уравнение 9ctg^2+4sin^2=6

9cos^2 x/sin^2 x + 4sin^2 x - 6 = 0 ОДЗ: sin x =/= 0, x =/= pi*k 9cos^2 x + 4sin^4 x - 6sin^2 x = 0 4sin^4 x - 6sin^2 x + 9 - 9sin^2 x = 0 4sin^4 x - 15sin^2 x + 9 = 0 sin^2 x = t, 0 <= t <= 1 при любом х 4t^2 - 15t + 9 = 0 D = 225 - 4*4*9 = 225 - 144 = 81 = 9^2 t1 = (15 - 9)/8 = 6/8 = 3/4 t2 = (15 + 9)/8 = 24/8 > 1 - не подходит sin^2 x = 3/4 Известно, что cos 2x = 1 - 2sin^2 x sin^2 x = (1 - cos 2x)/2 = 3/4 1 - cos 2x = 3/2 cos 2x = -1/2 2x1 = 2pi/3 + 2pi*k, x1 = pi/3 + pi*k 2x2 = 4pi/3 + 2pi*k, x2 = 2pi/3 + pi*k