How to find the value of x if #sec x = -5#?

1 Answer
Sep 17, 2015

#secx = -5#

We rewrite the secant function and do some algebric manipulation.

#1/cosx = -5#
#1 = -5cosx#
#-1/5 = cosx#
#cosx = -1/5#

We take the arccosine of both sides

#arccos(cosx) = arccos(-1/5)#

And evaluate that on a calculator, it comes out to approximately #101.5º#

The arccosine has a proper range so it outputs every value uniquely (We know it only outputs values on the range of #0 <= theta <= 180º#

But since we know that #cos(x+360) = cosx# we can further rewrite the answer to #S ~= 101.5º + 360nº#

Where #n# is an integer.

And that isn't all done, since there is another angle, this time on the third quadrant with the same cosine, we know that #cos(360-x) = cos(x)# and that #cos(-x) = cos(x)#

Therefore, #cos(360-101.5) = cos(101.5)#, so #cos(258.5) ~= -1/5# too.

Which leaves us
#S ~= 101.5º + 360nº and 258.5º + 360nº# where #n# is an integer.