Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

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

⇒ (sin x + sin 3x) – 3sin 2x – (cos x + cos 3x) + 3 cox 2x = 0

∵ sin A + sin B =

∴

⇒ 2 sin 2x cos x – 3 sin 2x – 2 cos 2x cos x + 3 cos 2x = 0

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

⇒ (2cos x – 3)(sin 2x – cos 2x) = 0

∴ cos x = 3/2 = 1.5 (not accepted as cos x lies between – 1 and 1)

Or sin 2x = cos 2x

∴ tan 2x = 1 = tan π/4

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴ 2x = nπ + π/4

∴