How to find the asymptotes of #y = 1/2 sec ( x - pi/6)#?

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Answer 1 · 2017-08-12T09:27:01

#x=2kpi+(3pi)/2 and x=2kpi-pi/3 , kinZZ#

#y=1/2sec(x-pi/6)=1/2*1/cos(x-pi/6)#

We know that when #cos(x-pi/6)rarr0# then #yrarrpmoo#
so everytime that #cos(x-pi/6)=0# the function has a vertical asymptode.

Let's solve the equation #cos(x-pi/6)=0# :

#cos(x-pi/6)=0iffcos(x-pi/6)=cos(pi/2)iff#

#x-pi/6=2kpipmpi/2iff x=2kpi+(2pi)/3or x=2kpi-pi/3#

So the function has the folowing vertical asyptodes :

#x=2kpi+(3pi)/2 and x=2kpi-pi/3 , kinZZ#