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Question

Solve the following equation:
2sin2xsin4xcos2x=sin3x

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Solution

cos(4x2x)cos(tan2x)cos2x=sin3x
(12sin23x)=sin3x
2sin23xsin3x1=0
(2sin3x+1)(sin3x1)=0
So, sin3x=12,3x=nπ+(1)n(π6),x=nπ3+(1)nπ18. and sin3x=1,3x=2nππ2,x=2nπ3π6.

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