For what values of x, if any, does #f(x) = sec((-11pi)/6-7x) # have vertical asymptotes?

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Answer 1 · 2017-12-16T08:59:43

#=> x = (-2pin) /7 pm -pi/14 -(11pi)/42 #

# n in ZZ #

We need to consider the definition of:

#sec x = 1/cosx #

Hence:

#sec( (-11pi)/6 -7x ) = 1/( cos( (-11pi)/6 -7x)) #

Hence there are verticle asymtptotes where the denominator #=0#

#=> cos( (-11pi)/6 - 7x) = 0 #

#=> cos ((-11pi)/6 - 7x ) = cos(pi/2) #

Using the general solution for #cosx#:

If #cosx = cos phi #
#=> x = 2pin pm phi #

#=>(-11pi)/6 - 7x = 2pin pm pi/2 #

#=> -7x = 2pin pm pi/2 + (11pi)/6 #

#=> x = (-2pin) /7 pm -pi/14 -(11pi)/42 #