Find the exact radian value of $\displaystyle Arccos(\frac{-\sqrt{3}}{2})$
My answers: $\displaystyle \theta = \frac{5\pi}{6} + 2k\pi, \frac{7\pi}{6} + 2k
\pi$
is this right?
Here are the two non-coterminal pairs.
$\displaystyle (\pi-\pi/6)+2\pi k$
$\displaystyle (\pi+\pi/6)+2\pi k$
Which simplifies to,
$\displaystyle -\pi/6+\pi(2k+1)$
$\displaystyle \pi/6+\pi(2k+1)$
Thus, all solutions are,
$\displaystyle \pm\frac{\pi}{6}+\pi k$ where $\displaystyle k\in\{...-5,-3,-1,1,3,5,...\}$
wait a second. the range of arccos is $\displaystyle [0,\pi]$ since it is the inverse of cos. This means that arccos is only negative in II. so the only answer would be $\displaystyle \frac{5\pi}{6}$. is my reasoning correct? i was just thinking about the graph of arccos.
Technically you are correct here. However, in all cases that I have seen, when you are asked to solve this kind of equation the solution set is assumed to be over the whole real line, not just the range given by arccos.Originally Posted by c_323_h
-Dan