Question

How do you solve `(1+cosx)/sinx + sinx/(1+cosx) = 4`?

Answer 1

First of all: the existence conditions.

`sinx!=0rArrx!=kpi`

and

`1+cosx!=0rArrcosx!=-1rArrx!=pi+2kpi`

So, together: `x!=kpi`.

Now:

`((1+cosx)^2+sin^2x)/(sinx(1+cosx))=4*(sinx(1+cosx))/(sinx(1+cosx))rArr`

`1+2cosx+cos^2x+sin^2x=4sinx+4sinxcosxrArr`

`2+2cosx-4sinx-4sinxcosx=0rArr`

`2(1+cosx)-4sinx(1+cosx)=0rArr`

`2(1+cosx)(1-2sinx)=0`.

So:

`1+cosx=0rArrcosx=-1rArrx=pi+2kpi` not acceptable,

and:

`1-2sinx=0rArrsinx=1/2rArr`

`x=pi/6+2kpi` acceptable

`x=5/6pi+2kpi` acceptable.