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

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Answer 1 · 2015-04-16T14:52:12

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.

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