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

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Answer 1 · 2016-10-04T07:37:30

In the interval #0<=x<=2pi#,

#x=pi/3# or #x=(5pi)/3#

#(1+sinx)/cosx+cosx/(1+sinx)#

= #(1+sinx)/cosx+(cosx(1-sinx))/((1+sinx)(1-sinx))#

= #(1+sinx)/cosx+(cosx-sinxcosx)/(1-sin^2x)#

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

= #(cosx+sinxcosx+cosx-sinxcosx)/cos^2x#

= #(2cosx)/cos^2x=2/cosx=2secx#

Hence, #(1+sinx)/cosx+cosx/(1+sinx)=4#

#hArr2secx=4# or #secx=2#

As #sec(+-pi/3)=2#

and in the interval #0<=x<=2pi#, #x=pi/3# or #x=2pi-pi/3=(5pi)/3#