For what values of x, if any, does #f(x) = cot((7pi)/12-x) # have vertical asymptotes?

Question

1429 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-31T20:49:36

#x= (7pi)/12 - 2pin #
#x = -(5pi)/12 - 2pin#

For #n in ZZ#

To approach this we must cosnider how #cotx = cosx/sinx#

Hence for #cotx# to have verticle asymptotes, the denominator must be 0, so hence #sinx = 0#

So hence in this circumstance ;
#sin((7pi)/12 - x) = 0# for the verticle asymptotes

So hence now we consider the general solution of;
#sinx = sina#
Then #x = 2pin + a#
and #x = 2pin + pi -a#

So hence #sin( (7pi) /12 - x) = sin0#

Hence; #(7pi)/12 - x = 2pin#
and #(7pi)/12 - x =2pin + pi#

Hence rearanging to yeild;

#x= (7pi)/12 - 2pin #
#x = -(5pi)/12 - 2pin#

For #n in ZZ#

We can consider this by graphing;
graph{cot( 7pi/12 -x) [-4.354, 5.644, -2.28, 2.72]}