The general solution of sin 3x + sin x – 3 sin 2x = cos 3x + cos x �

1.	The general solution of sin 3x + sin x – 3 sin 2x = cos 3x + cos x – 3 cos 2x is(a) \(\frac{nπ}{2}+ \frac{π}{8}\) , n ∈ I(b) \(\frac{nπ}{2}- \frac{π}{8}\) , n ∈ I (c) \(nπ+ \frac{π}{8}\) , n ∈ I(d) \(nπ- \frac{π}{8}\) , n ∈ I
Answer» Answer : (a) \(\frac{nπ}{2}+\frac{π}{8}\), n ∈ I (sin 3x + sin x) – 3 sin 2x = (cos 3x + cos x) – 3 cos 2x ⇒ 2 sin 2x cos x – 3 sin 2x = 2 cos 2x . cos x – 3 cos 2x ⇒ sin 2x (2 cos x – 3) = cos 2x (2 cos x – 3) ⇒ (2 cos x – 3) (sin 2x – cos 2x) = 0 ⇒ (2 cos x – 3) = 0 or (sin 2x – cos 2x) = 0 ⇒ cos x = \(\frac{3}{2}\) or sin 2x = cos 2x ∵ – 1 ≤ cos x ≤ 1, cos x ≠ \(\frac{3}{2}\) ∴ sin 2x = cos 2x ⇒ tan 2x = 1 ⇒ tan 2x = tan (π/4) ⇒ 2x = nπ + π/4 ⇒ x = \(\frac{nπ}{2}+\frac{π}{8}\), n ∈ I

The general solution of sin 3x + sin x – 3 sin 2x = cos 3x + cos x – 3 cos 2x is(a) \(\frac{nπ}{2}+ \frac{π}{8}\) , n ∈ I(b) \(\frac{nπ}{2}- \frac{π}{8}\) , n ∈ I (c) \(nπ+ \frac{π}{8}\) , n ∈ I(d) \(nπ- \frac{π}{8}\) , n ∈ I

Answer»

Answer : (a) \(\frac{nπ}{2}+\frac{π}{8}\), n ∈ I

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

⇒ 2 sin 2x cos x – 3 sin 2x = 2 cos 2x . cos x – 3 cos 2x

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

⇒ (2 cos x – 3) (sin 2x – cos 2x) = 0

⇒ (2 cos x – 3) = 0 or (sin 2x – cos 2x) = 0

⇒ cos x = \(\frac{3}{2}\) or sin 2x = cos 2x

∵ – 1 ≤ cos x ≤ 1, cos x ≠ \(\frac{3}{2}\)

∴ sin 2x = cos 2x ⇒ tan 2x = 1

⇒ tan 2x = tan (π/4)

⇒ 2x = nπ + π/4

⇒ x = \(\frac{nπ}{2}+\frac{π}{8}\), n ∈ I