Here,
#((cosx+1)/sinx)^2=1/3...to(A)#
#=>(cos^2x+2cosx+1)/sin^2x=1/3#
#=>3cos^2x+6cosx+3=sin^2x#
#=>3cos^2x+6cosx+3=1-cos^2x#
#=>4cos^2x+6cosx+2=0#
#=>2cos^2x+3cosx+1=0#
#=>2cos^2x+2cosx+cosx+1=0#
#=>2cosx(cosx+1)+1(cosx+1)=0#
#=>(cosx+1)(2cosx+1)=0#
#=>cosx+1=0 or 2cosx+1=0#
#=>cosx=-1 orcosx=-1/2#
#(1)cosx=-1# ,does not satify #(A)#
#i.e. cosx=-1 # is rejected.
#(2)cosx=-1/2=>cosx=cos((2pi)/3)#
#=>color(red)(x=2kpi+-(2pi)/3 ,k inZZ#
Note:
#(1)# If #cosx=-1# ,then #sinx=0=>#Point #A'(-1,0)# on unit circle .
#=>LHS=((cosx+1)/sinx)^2=((-1+1)/0)=oo!=1/3#
#color(red)((2)costheta=cosalpha=>theta=2kpi+-alpha,kinZZ#