Need double checking.

**c_323_h** · March 26th 2006, 01:57 PM

Find the exact radian value of $Arccos(\frac{-\sqrt{3}}{2})$

My answers: $\theta = \frac{5\pi}{6} + 2k\pi, \frac{7\pi}{6} + 2k<br /> \pi$

is this right?

**ThePerfectHacker** · March 26th 2006, 02:02 PM

Originally Posted by c_323_h

Find the exact radian value of $cos^-1(\frac-{\sqrt3}{2})$

My answers: $\theta = \frac{5\pi}{6} + 2k\pi and \frac{7\pi}{6} + 2k\<br /> pi$

is the right?

Notice that,
$-\frac{\sqrt{3}}{2}$ to get that you need,
$x=\frac{5\pi}{6}$ thus, all the solutions are,
$x=\frac{5\pi}{6}+\pi k,k\in\mathbb{Z}$

**c_323_h** · March 26th 2006, 02:15 PM

Originally Posted by ThePerfectHacker

Notice that,
$-\frac{\sqrt{3}}{2}$ to get that you need,
$x=\frac{5\pi}{6}$ thus, all the solutions are,
$x=\frac{5\pi}{6}+\pi k,k\in\mathbb{Z}$

why pi and not 2pi?

**ThePerfectHacker** · March 26th 2006, 03:40 PM

Originally Posted by c_323_h

why pi and not 2pi?

Your answers are right, by my answer is simpler.

**topsquark** · March 26th 2006, 03:48 PM

Originally Posted by ThePerfectHacker

Your answers are right, by my answer is simpler.

ThePerfectHacker, $cos(5 \pi/6 + \pi) = + \sqrt3/2$ so your form for the solution is not correct.

Your original answer is correct, c_323_h.

-Dan

**ThePerfectHacker** · March 26th 2006, 04:20 PM

Here are the two non-coterminal pairs.
$(\pi-\pi/6)+2\pi k$
$(\pi+\pi/6)+2\pi k$
Which simplifies to,
$-\pi/6+\pi(2k+1)$
$\pi/6+\pi(2k+1)$
Thus, all solutions are,
$\pm\frac{\pi}{6}+\pi k$ where $k\in\{...-5,-3,-1,1,3,5,...\}$

**c_323_h** · March 26th 2006, 04:43 PM

Originally Posted by ThePerfectHacker

Here are the two non-coterminal pairs.
$(\pi-\pi/6)+2\pi k$
$(\pi+\pi/6)+2\pi k$
Which simplifies to,
$-\pi/6+\pi(2k+1)$
$\pi/6+\pi(2k+1)$
Thus, all solutions are,
$\pm\frac{\pi}{6}+\pi k$ where $k\in\{...-5,-3,-1,1,3,5,...\}$

what does $\in$ mean in layman's terms?

**ThePerfectHacker** · March 26th 2006, 05:58 PM

Originally Posted by c_323_h

what does $\in$ mean in layman's terms?

It means "element of"

That means $k$ must be one of the "elements" of that set.

**c_323_h** · March 26th 2006, 06:05 PM

wait a second. the range of arccos is $[0,\pi]$ since it is the inverse of cos. This means that arccos is only negative in II. so the only answer would be $\frac{5\pi}{6}$ . is my reasoning correct? i was just thinking about the graph of arccos.

**topsquark** · March 27th 2006, 03:57 AM

Originally Posted by c_323_h

wait a second. the range of arccos is $[0,\pi]$ since it is the inverse of cos. This means that arccos is only negative in II. so the only answer would be $\frac{5\pi}{6}$ . 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.

-Dan

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