What is arcsec (-2)? The question asks to solve $\operatorname{arcsec}(-2) $.

Options are

a) &pi;/3

b) &minus;(&pi;/3)

c) 5&pi;/3

d) 4&pi;/3

e) 2&pi;/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&deg;. In quadrant 2 that is 2&pi;/3. In quadrant 3 it's 4&pi;/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?
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&lt;\frac{\pi}{2} &amp; \text{if $x\ge1$}\\
\frac{\pi}{2}&lt;\operatorname{arcsec}x\le\pi &amp; \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.
The cos -.5 is either 120 degrees in Q II or 240 degrees in Q III.  both answers are correct!!