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.