Sinx+sin3x-sin4x=0 решить уравнение

sinx + sin3x - sin4x = 0,

2 * (sin ((x + 3 х) / 2) * cos ((3x - x) / 2) - sin4x = 0,

2 * sin2x * cosx - sin4x = 0,

2 * sin2x * cosx - (2 * sin4x/2 * cos4x/2) = 0,

2 * sin2x * cosx - 2sin2x * cos2x = 0,

2 * sin2x * (cosx - cos2x) = 0,

1) sin2x = 0,

x = + / - arcsin0 + 2 пn, n э z, x = пn, n э z,

2) cosx - cos2x = 0,

cos2x - cosx = 0,

- 2 * sin3x/2 * sinx/2 = 0,

sin (3x/2) = 0,

x = + / - arcsin0 + пn * 2/3, n э z, x = 2/3 пn, n э z.

3) sin (x/2) = 0,

x = arcsin0 + пn * 2, n э z, x = 2 пn, n э z.