For what values of x, if any, does #f(x) = sec((-7pi)/4-2x) # have vertical asymptotes?

1 Answer
Nov 7, 2017

Answer:

#x=((-9pi)/8)-(pik)/2# Where #k=...-2,-1,0,1,2,...#

Explanation:

The first thing to do in this problem is recognize a trig identity. Notably, #sec(x)=1/cos(x)#. That means we can rewrite what you have as #1/cos((-7pi)/4-2x)#.

To find the value of a vertical asymptote, you must set the denominator equal to zero and solve. So the set up in this case is: #cos((-7pi)/4-2x)=0#. What follows from here on out is algebra and trig operations.

We begin by performing the #arccos(x)# operation on both sides: #arccos(cos((-7pi)/4-2x))=arccos(0)#. This eliminates the #cos# from the left side and for the right side we must call upon our memory of the unit circle. We can think of the right side as asking the question "at what angle on the unit circle does #cos(x)=0#?"

The answer to that question gives us #pi/2#. However, we know that if we add #pi# to that value, we arrive at another place on the unit circle where #cos(x)=0#, so to symbolize that, we add #pik# where #k=...-2,-1,0,1,2...#, that is, any whole number integer. So the equation looks like #(-7pi)/4-2x=pi/2+(pik)#.

To finish, we add #(7pi)/4# to both sides and after, divide both sides by #-2#. This results in #x=(-pi)/4-(7pi)/8-(pik)/2# where #k=...-2,-2,0,2,1...#. If we want to simplify, we get #x=(-9pi)/8-(pik)/2# where #k=...-2,-1,0,1,2...#