Q) Solve the following equation;
sinx-3sin2x+sin3x=cosx-3cos2x+cos3x.
Asked by Anish
9th September 2018,
8:02 PM
Answered by Expert
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
Answered by Expert
10th September 2018,
9:27 AM
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