How do you find all solutions of the equations #secx+tanx=1# in the interval #[0,2pi)#?
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The solutions are #S={0}# for #x in [0, 2pi)#
We need
#secx=1/cosx#
#tanx=sinx/cosx#
Therefore,
#secx+tanx=1#
#1/cosx+sinx/cosx=1#
Simplifying
1+sinx=cosx#
#cosx-sinx=1#
We will apply
#rcos(x+a)=r(cosxcosa-sinxsina)#
#rcosa=1#
#rsina=1#
#r^2(sin^2a+cos^2a)=1+1=2#
Therefore,
#r^2=2#, #=>#, #r=sqrt2#
#cosa=1/sqrt2#, #=>#, #a=1/4pi#
#sina=1/sqrt2#, #=>#, #a=1/4pi#
#sqrt2(1/sqrt2cosx-1/sqrt2cosx)=1#
#cos(x+1/4pi)=1/sqrt2#
#x+1/4pi=1/4pi+2kpi# and #x+1/4pi=-1/4pi+2kpi#
#x=2kpi# and #x=-1/2pi+2kpi#
When #x=3/2pi#, #=>#, #x in O/#
Given that, #secx+tanx=1......................................(1).#
We know that, #sec^2x-tan^2x=1, i.e.,#
#(secx+tanx)(secx-tanx)=1.#
#:. (1)(secx-tanx)=1..................................[because, (1)],#
#:. secx-tanx=1............................................................(2).#
#:. (1)+(2) rArr 2secx=2, or, secx=1.#
#rArr cosx=1=cos0.........................................................(3).#
Since, # costheta=cosalpha rArr theta=2kpi+-alpha, k in ZZ,# we have,
from #(3), x =2kpi, k in ZZ.#
Hence, the reqd. soln. in #[0,2pi)# is, #x=0.#
