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

1 Answer
Jun 12, 2018

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

Explanation:

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