How to solve #((cosx+1)/sinx)^2=1/3#?

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Answer 1 · 2018-06-12T10:45:45

#=>x=2kpi+-(2pi)/3 ,k inZZ#

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#