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 #