Question

Solve sin x-3 sin 2x + sin 3x = COS X - 3 cos 2x + cos3x​

Answer 1

Answer:

sinx - 3sin2x + sin3x = cosx - 3cos2x + cos3x

sinx + sin3x - 3sin2x = cosx + cos3x - 3cos2x

2sin2xcosx - 3 sin2x - 2cos2x cosx + 3cos2x = 0

sin2x(2cosx - 3) - cos2x(2cosx - 3) = 0

(2cosx - 3)(sin2x - cos2x) = 0

sin2x = cos2x as cos x ≠ 3/2

2x = 2n∏ ± (∏/2 - 2x)

x = n∏/2 + ∏/8