The general solution of the equation `8cosxcos2xcos4x=(sin6x)/(sin x)` is `x=((npi)/7)+(pi/(21)),AAn in Z` `x=((2pi)/7)+(pi/(14)),AAn in Z` `x=((npi)/7)+(pi/(14)),AAn in Z` `x=(npi)+(pi/(14)),AAn in Z`

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1 Answers

`4(2sinxcosx)cos2xcos4x=sin6x`
`4sin2xcos2xcos4x=sin6x`
`2(2sinxcos2x)cos4x=sin6x`
`2sin4xcos4x=sin6x`
`sin8x-sin6x=0`
`2cos((8x+6x)/2)*sin((8x-6x)/2)=0`
`2cos7xsinx=0`
`cos7x=0`
`7x=npi+pi/2`
`x=npi/7+pi/14`
Option B is correct.

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