How do you find all solutions of the equations #secx+tanx=1# in the interval #[0,2pi)#?

2 Answers
Narad T.
Nov 13, 2017

The solutions are #S={0}# for #x in [0, 2pi)#

Explanation:

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/#

Ratnaker Mehta
Nov 13, 2017

# x=0.#

Explanation:

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.#

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