How do you solve #cosx/(1+sinx)+(1+sinx)/cosx=-4# for #0<=x<=2pi#?

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Answer 1 · 2016-11-26T06:24:20

#x=(2pi)/3# or #(4pi)/3#

We have #cosx/(1+sinx)+(1+sinx)/cosx=-4#.

This can be written as

#(cos^2x+(1+sinx)^2)/(cosx(1+sinx))=-4#

or #(cos^2x+1+sin^2x+2sinx)/(cosx(1+sinx))=-4#

or #(1+1+2sinx)/(cosx(1+sinx))=-4#

or #(2(1+sinx))/(cosx(1+sinx))=-4#

or #2/cosx=-4#

or #cosx=-2/4=-1/2=cos(+-(2pi)/3)#

i.e. #x=2npi+-(2pi)/3#, where #n# is an integer

and if #0<=x<=2pi#,

#x=(2pi)/3# or #(4pi)/3#