Find the general solution of the equation sin x – 3sin2x + sin3x = cos x – 3cos2x + cos3x

Given,

sin x – 3sin2x + sin3x = cos x – 3cos2x + cos3x

To solve the problem we need to apply transformation formula, but we need to group sin x and sin 3x together and similarly cos x and cos 3x in RHS side.

∴ sin x + sin3x – 3sin2x = cos x + cos3x – 3cos2x

Transformation formula:

cos A + cos B =

sin A + sin B =

⇒

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

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

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

⇒ (sin 2x – cos 2x)(2cos x – 3) = 0

⇒ cos x = 3/2 or sin 2x = cos 2x

As cos x ∈ [-1,1]

∴ no value of x exists for which cos x = 3/2

∴ sin 2x = cos 2x

⇒ tan 2x = 1 = tan π/4

We know solution of tan x = tan α is given by –

x= nπ + α , n ∈ Z

∴ 2x = nπ + (π/4)

⇒