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How do you solve #cos^3xsin^2x=(sin^2x-sin^4x)cosx#?

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Jun 25, 2017

#LHS=cos^3xsin^2x#

#=cosxcos^2xsin^2x#

#=cosx(1-sin^2x)sin^2x#

#=cosx(sin^2x-sin^4x)=RHS#

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