According to the question,
sin x – 3sin2x + sin3x = cos x – 3cos2x + cos3x
Grouping sin x and sin 3x in LHS and, cos x and cos 3x in RHS,
We get,
sin x + sin3x – 3sin2x = cos x + cos3x – 3cos2x
Applying transformation formula,
cos A + cos B = 2cos ((A + B)/2) cos((A – B)/2)
sin A + sin B = 2sin ((A + B)/2) cos((A – B)/2)
⇒
⇒ 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]
Hence, no value of x exists for which cos x = 3/2
Therefore, 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
Therefore, 2x = nπ + (π/4)
⇒ x = nπ/2 + (π/8), n ∈ Z
