College Trigonometry, 2011a

840 Foundations of Trigonometry
to tan(θ) =−2 are coterminal with β. Moreover, the solutions from Quadrant II differ by
exactly π units from the solutions in Quadrant IV, so all the solutions to tan(θ) =−2 areof
the form θ = β + πk =arctan(−2) + πk for some integer k. Switching back to the variable t,
we record our final answer to tan(t) =−2 ast =arctan(−2) + πk for integers k.
3. The last equation we are asked to solve, sec(x) =− 5 3
, poses two immediate problems. First,
we are not told whether or not x represents an angle or a real number. We assume the latter,
but note that we will use angles and the Unit Circle to solve the equation regardless. Second,
as we have mentioned, there is no universally accepted range of the arcsecant function. For
that reason, we adopt the advice given in Section 10.3 and convert this to the cosine problem
cos(x) =− 3 5 . Adopting an angle approach, we consider the equation cos(θ) =− 3 5
and note
that our solutions lie in Quadrants II and III. Since − 3 5
isn’t the cosine of any of the ‘common
angles’, we’ll need to express our solutions in terms of the arccosine function. The real number
t = arccos ( − 3 5)
is defined so that
π
2