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Question

If 8sinθcosθcos2θcos4θ=sinx, then x=
  1. 32θ
  2. 16θ
  3. 8θ
  4. 4θ


Solution

The correct option is C 8θ
8sinθcosθcos2θcos4θ=2[2(2sinθcosθ)cos2θ]cos4θ=2[2sin2θcos2θ]cos4θ=2sin4θcos4θ=sin8θx=8θ

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