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 3x – sin x = 4 cos² x – 2

⇒ sin 3x – sin x = 2(2 cos² x – 1)

⇒ sin 3x – sin x = 2 cos 2x {∵ cos 2θ = 2cos² θ – 1}

{∵ sin A - sin B =

⇒

⇒

⇒

∴ either, cos 2x = 0 or sin x = 1 = sin π/2

In case of cos x = 0 we can give solution directly as cos x = 0 is true for x = odd multiple of π/2

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

∴

⇒ where m, n ϵ Z