How do you solve `(cot x)+ sqrt(3) = (csc x)`?

Identities:

`color(red)bb(cotx=cosx/sinx)`

`color(red)bb(cscx=1/sinx)`

Substituting these into given equation:

`cosx/sinx+sqrt(3)=1/sinx`

`cosx/sinx+sqrt(3)-1/sinx=0`

Add LHS:

`(cosx+sqrt(3)sinx-1)/sinx=0`

This can only be true if the numerator equals zero. we can't have division by zero.

`:.`

`cosx+sqrt(3)sinx-1=0`

`cosx-1=-sqrt(3)sinx`

Square both sides:

`(cosx-1)^2=3sin^2x`

Identity:

`color(red)bb(sin^2x+cos^2x=1)`

`(cosx-1)^2=3(1-cos^2x)`

Expanding:

`cos^2x-2cosx+1=3-3cos^2x`

`4cos^2x-2cosx-2=0`

Let `m=cosx`

`:.`

`4m^2-2m-2=0`

Factor:

`(2m+1)(2m-2)=0=>m=-1/2 and m=1`

`x=arccos(cosx)=arccos(-1/2)=>color(blue)(x=(2pi)/3+n2pi)`

`x=arccos(cosx)=arccos(1)=>x=2pin`

`cscx` and `cotx` are undefined for `2pin` (division by zero)

Only:

`color(blue)(x=(2pi)/3+2npi)`

`n in ZZ`

Here,

`cotx+sqrt3=cscx`

`=>cosx/sinx+sqrt3=1/sinx`

We take , denominator `color( blue)(sinx!=0=>x!=kpi,kinZZ.`

So,

`cosx+sqrt3sinx=1`

Dividing both sides by `2`

`1/2cosx+sqrt3/2sinx=1/2`

`=>cosxcos(pi/3)+sinxsin(pi/3)=cos(pi/3)...to color(green) (Apply(1)`

`=>cos(x-pi/3)=cos(pi/3)`

`=>x-pi/3=2kpi+-pi/3,kinZZ...tocolor(green)( Apply(2)`

`=>x=2kpi+pi/3+pi/3,kinZZ or x=2kpi+pi/3-pi/3,kinZZ`

`=>x=2kpi+(2pi)/3,kinZZ or x=2kpi,kinZZ`

But , `x!=kpi, kinZZ=>x!=2kpi,kinZZ`

Hence,

`=x=2kpi+(2pi)/3,kinZZ`