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 $cosx = cos phi$
$=> x = 2pin pm phi$

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

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

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

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