(sin 3x + sin x) / (cos 3x + cos x) = sqrt 3, sin x = ?

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Answer 1 · 2018-02-25T15:39:14

$sin x=1/2$ [when $0<=x<=pi/2$ ]

$(sin 3x + sin x) / (cos 3x + cos x) = sqrt 3$

We'll apply 2 identities here, $color(red)(sin3x = 3sinx-4sin^3x)$ and $color(red)(cos3x=4cos^3x-3cosx)$

or, $(color(red)(3sinx-4sin^3x)+sinx)/(color(red)(4cos^3x-3cosx)+cosx)=sqrt3$

Combining $sinx$ terms,

or, $(4sinx-4sin^3x)/(4cos^3x-2cosx)= sqrt3$

or, $[4sinx(1-sin^2x)]/[2cosx(2cos^2x-1)]=sqrt3$

Again, 2 more identities, $color(magenta)(cos^2x = 1-sin^2x$ and $color(magenta)(cos2x=2cos^2x-1$

or, $(2sinxcancel(color(magenta)(cos^2x))^(cosx))/(cancelcosxcolor(magenta)(cos2x))= sqrt3$

or, $(2sinxcosx)/(cos2x)=sqrt3$

One more here, $color(blue)(sin2x=2sinxcosx$

or, $color(blue)(sin2x)/(cos2x)= sqrt3$

or, $tan2x= sqrt3$

or, $tan2x= tan(pi/3)$

or, $2x= pi/3$

or, $x= pi/6$

so, $sinx= sin(pi/6)= 1/2$

Answer 2 · 2018-02-25T16:27:02

$sinx=1/2$

We may use,

$rarrsinA+sinB=2sin((A+B)/2)*cos((A+B)/2)$ and

$rarrcosA+cosB=2cos((A+B)/2)*cos((A+B)/2)$

Given that,

$rarr(sin3x+sinx)/(cos3x+cos)=sqrt(3)$

$rarr(cancel(2)sin((3x+x)/2)*cancel(cos((3x-x)/2)))/(cancel(2)cos((3x+x)/2)*cancel(cos((3x-x)/2)))=sqrt(3)$

$rarrtan2x=tan60°$

$rarr2x=60°$

$rarrx=30°$

Now, $sinx=sin30°=1/2$