Question

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

Answer 1

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

Explanation:

`(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

`sinx=1/2`

Explanation:

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`