Solution :
we have, cos3x + (sin2x – sin4x) = 0
= cos3x – 2sinx.cos3x = 0
⟹ (cos3x)(1 – 2sinx) = 0
⟹ cos3x = 0 or sinx = 12
⟹ cos3x = 0 = cosπ2 or sinx = 12 = sinπ6
⟹ 3x = 2nπ ± π2 or x = mπ + (−1)mπ6
⟹ x = 2nπ3 ± π6 or x = mπ + (−1)mπ6; (n, m ∈ I)
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