Find the general solution of sin x + sin 3x + sin 5x = 0.

1.	Find the general solution of sin x + sin 3x + sin 5x = 0.
Answer» sin x + sin 3x + sin 5x = 0 ∴ (sin 5x + sin x) + sin 3x = 0 ∴ 2sin \((\frac{5x+x}{2})\) cos \((\frac{5x-x}{2})\) + sin 3x = 0 ∴ 2sin 3x cos 2x + sin 3x = 0 ∴ sin 3x (2cos 2x + 1) = 0 ∴ sin 3x = 0 or 2cos 2x + 1 = 0 ∴ sin 3x = 0 or 2cos 2x = \(-\frac{1}{2}\) ∴ sin 3x = 0 or cos 2x = - cos \(\frac{π}{3}\) = cos \((π-\frac{π}{3})\) ∴ sin 3x = 0 or cos 2x = cos \(\frac{2π}{3}\) Since, sin θ = 0 implies θ = nπ and cos θ = cos α implies θ = 2nπ ± α , n ∈ Z. ∴ 3x = nπ or 2x = 2mπ ± \(\frac{2π}{3}\), ∴ the required general solution is x = \(\frac{nπ}{3}\) or x = mπ ± \(\frac{π}{3}\) where n, m ∈ Z.

Answer»

sin x + sin 3x + sin 5x = 0

∴ (sin 5x + sin x) + sin 3x = 0

∴ 2sin \((\frac{5x+x}{2})\) cos \((\frac{5x-x}{2})\) + sin 3x = 0

∴ 2sin 3x cos 2x + sin 3x = 0

∴ sin 3x (2cos 2x + 1) = 0

∴ sin 3x = 0 or 2cos 2x + 1 = 0

∴ sin 3x = 0 or 2cos 2x = \(-\frac{1}{2}\)

∴ sin 3x = 0 or cos 2x = - cos \(\frac{π}{3}\) = cos \((π-\frac{π}{3})\)

∴ sin 3x = 0 or cos 2x = cos \(\frac{2π}{3}\)

Since, sin θ = 0 implies θ = nπ and cos θ = cos α implies θ = 2nπ ± α , n ∈ Z.

∴ 3x = nπ or 2x = 2mπ ± \(\frac{2π}{3}\),

∴ the required general solution is x = \(\frac{nπ}{3}\) or x = mπ ± \(\frac{π}{3}\) where n, m ∈ Z.

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