The general solution of sin x - 3sin 2x + sin 3x = cos x - 3 cos 2x + cos 3x.

Solution:

Given sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x.

We know sin C + sin D = 2 sin (C+D)/2 cos (C-D)/2

Also cos C + cos D = 2 cos (C+D)/2 cos (C-D)/2

(sin x + sin 3x) – 3sin 2x = (cos x + cos 3x) – cos 2x

=> 2sin 2x cos x – 3sin 2x = 2cos 2x cos x – 3cos 2x

sin 2x(2 cos x – 3) = cos 2x(2 cos x – 3)

Cancel (2 cos x – 3) from both sides

=> sin 2x = cos 2x

=> tan 2x = 1

=> 2x = nπ + (π/4)

=> x = (nπ/2) + (π/8)

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