=> sin 6x = sin 4x – sin 2x
=> sin 6x + sin 2x = sin 4x
=> 2sin{(6x + 2x)/2}.cos{(6x - 2x)/2} - sin4x = 0
=> 2 sin 4x . cos 2x – sin 4x = 0
=> sin 4x[2 cos 2x – 1] = 0
=> sin 4x = 0 or 2 cos 2x – 1 = 0
=> 4x = nπ, n ∈ Z or cos 2x = 1/2 = cos π/3
=> x = nπ/4, n ∈ Z or 2x = 2nπ ± π/3, n ∈ Z
Solution set
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