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