What is arcsec (-2)? The question asks to solve $\operatorname{arcsec}(-2) $.
Options are
a) π/3
b) −(π/3)
c) 5π/3
d) 4π/3
e) 2π/3
So my thoughts are: this is quadrant 2 or 3; on the unit circle, $ x = -1/2 $.
$\cos(t) = 1/2$ where $t$ is 60°. In quadrant 2 that is 2π/3. In quadrant 3 it's 4π/pi.
I though that the answer could be either d or e.
The answer key says that the only right answer is d.
The graphing calculator shows that the answer must be e.
Who is right and why?
 A: It depends on how you define $\operatorname{arcsec} x$. Its domain is $(-\infty,-1]\cup[1,\infty)$, and one can define it so that
\begin{cases}
0\le \operatorname{arcsec}x<\frac{\pi}{2} & \text{if $x\ge1$}\\
\frac{\pi}{2}<\operatorname{arcsec}x\le\pi & \text{if $x\le-1$}
\end{cases}
You're in the second case, so (e) should be the correct answer, because $\cos(2\pi/3)=-1/2$.
However a different definition of the function is conceivable, although answer (d) would sound quite strange.
A: Wikipedia says "Some authors define the range of arcsecant to be ($0 \le y < 
\pi/2$ or $\pi \le y < 3\pi/2$), because the tangent function is nonnegative on this domain."
It is likely that your course adheres to this convention (and you are requested to comply).
A: Trig functions are not invertible -- trig functions do not have inverses. Thus, there are some problems involved with the idea of "inverse trig functions"!
The most familiar related example is square roots. The number 9 has two square roots. However, we've chosen one of those square roots to be the "principal" value, so when we write $\sqrt{9}$, we mean $3$.
We could have chosen the principal value to always be negative, so that we'd say $\sqrt{9} = -3$ and $\sqrt{9} \neq 3$. We could have even chosen it to be positive for some values and negative for others. But, in some sense, taking the positive square root is the 'right' convention.
The same is done for the inverse trig functions. This is why there is only one answer to your question rather than two. But unlike square root, there isn't a 'right' convention for some of them. Here's an excerpt from the Handbook of Mathematical Sciences, 6th Edition:

$$ \begin{array}
\\\hline
-\pi/2 &\leq& \mathop{\mathrm{Arcsin}} x &\leq& \pi/2, & & -1 &\leq& x &\leq& 1
\\\hline
0 &\leq& \mathop{\mathrm{Arccos}} x &\leq& \pi, & & -1 &\leq& x &\leq& 1
\\\hline
-\pi/2 &<& \mathop{\mathrm{Arctan}} x &<& \pi/2, & & -\infty &<& x &<& \infty
\\\hline
0 &<& \mathop{\mathrm{Arccsc}} x &\leq& \pi/2, & & & & x &\geq& 1
\\-\pi &<& \mathop{\mathrm{Arccsc}} x &\leq& -\pi/2, & & & & x &\leq& -1
\\\hline
0 &\leq& \mathop{\mathrm{Arcsec}} x &<& \pi/2, & & & & x &\geq& 1
\\ -\pi &\leq& \mathop{\mathrm{Arcsec}} x &<& -\pi/2, & & & & x &\leq& -1
\\\hline
0 &<& \mathop{\mathrm{Arccot}} x &<& \pi, & & -\infty &<& x &<& \infty
\\\hline
\end{array}$$
  Note. There is no uniform agreement on the definitions of $\mathop{\mathrm{Arccsc}} x$, $\mathop{\mathrm{Arcsec}} x$, $\mathop{\mathrm{Arccot}} x$ for negative values of $x$

which, as you can see, would have assigned a different value to $\mathop{\mathrm{Arcsec}} -2$ that doesn't even appear on your list of choices!
A: The cos -.5 is either 120 degrees in Q II or 240 degrees in Q III.  both answers are correct!!
