Question #787c0

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Answer 1 · 2016-09-25T09:58:05

$4sin4xsin2xsinx=sin3x$

$=>2*2sin4xsin2xsinx=sin3x$

$=>2(cos(4x-2x)-cos(4x+2x))sinx=sin3x$

$=>2cos(2x)sinx-2cos(6x))sinx=sin3x$

$=>sin(2x+x)-sin(2x-x)-(sin(6x+x)-sin(6x-x))=sin3x$

$=>sin3x-sinx-sin7x+sin5x=sin3x$

$=>sin7x-sin5x+sinx=0$

$=>2cos6xsinx+sinx=0$

$=>sinx(2cos6x+1)=0$

So $sinx=0$

$=>x=npi" where "ninZZ$

Again

$=>(2cos6x+1)=0$

$=>cos6x=-1/2=cos((2pi)/3)$

$=>6x=2npi+-(2pi)/3" where "n inZZ$

$=>x=(npi)/3+-(pi)/9" where "n inZZ$