Решите уравнение 4 sin^2 3x - sin 3x = 2 + sin^2 3x

sin^2 3x+3 cos^2 3x-4 sin (П/2+3 х) cos (П/2+3 х) = 0

sin^2 3x+3 (1 - sin^2 3x) - 2 * 2sin2 (П/2+3 х) = 0

sin^2 3x+3-3sin^2 3x - 4cos2x = 0

-2sin^2 3x - 4 (cos^2 x-sin^2 x) = 0

-2sin^2 3x - 4cos^2 x + 4sin^2x=0

2sin^2 x - 4 (1-sin^2x) = 0

2sin^2 x - 4 + 4sin^2 x=0

6 sin^2 x=4

sin^2x=4/6

sin^2 x=2/3

sinx=2/3 sinx=-2/3

x = (-1) ^n arcsin (2/3) + Пn x = (-1) ^n arcsin (-2/3) + Пn

x = (-1) ^n - arcsin (2/3) + Пn

x = (-1) ^n+1 arcsin (2/3) + Пn