Question #787c0

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Answer 1 · 2016-09-25T09:58:05

$4 sin 4 x sin 2 x sin x = sin 3 x$ $4sin4xsin2xsinx=sin3x$

$\Rightarrow 2 \cdot 2 sin 4 x sin 2 x sin x = sin 3 x$ $=>2*2sin4xsin2xsinx=sin3x$

$\Rightarrow 2 (cos (4 x - 2 x) - cos (4 x + 2 x)) sin x = sin 3 x$ $=>2(cos(4x-2x)-cos(4x+2x))sinx=sin3x$

$\Rightarrow 2 cos (2 x) sin x - 2 cos (6 x)) sin x = sin 3 x$ $=>2cos(2x)sinx-2cos(6x))sinx=sin3x$

$\Rightarrow sin (2 x + x) - sin (2 x - x) - (sin (6 x + x) - sin (6 x - x)) = sin 3 x$ $=>sin(2x+x)-sin(2x-x)-(sin(6x+x)-sin(6x-x))=sin3x$

$\Rightarrow sin 3 x - sin x - sin 7 x + sin 5 x = sin 3 x$ $=>sin3x-sinx-sin7x+sin5x=sin3x$

$\Rightarrow sin 7 x - sin 5 x + sin x = 0$ $=>sin7x-sin5x+sinx=0$

$\Rightarrow 2 cos 6 x sin x + sin x = 0$ $=>2cos6xsinx+sinx=0$

$\Rightarrow sin x (2 cos 6 x + 1) = 0$ $=>sinx(2cos6x+1)=0$

So $sin x = 0$ $sinx=0$

$=>x=npi" where "ninZZ$

Again

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

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

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

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