Question

solution help

Answer 1

How far did you get with it?

Answer 2

divided 24 on both sides

Answer 3

im stuck with this problem

Answer 4

Another way of saying \(\csc^2x\) is \((\csc x)^2\). Does that help?

Answer 5

That makes the equation: \((\csc x)^2=\frac{32}{24}\)

Answer 6

do i square root both sides?

Answer 7

Yes. And remember the \(\pm\) issues that causes.

Answer 8

how do you get the solutions once you square root it?

Answer 9

Well, that one does not look nice and friendly, so you would first flip it, then use arcsine in a calculator.

Answer 10

Hmmm... Or wait... that might eb nicer after being flipped.

Answer 11

yeah i got 42

Answer 12

\(\csc x=\sqrt{\frac{32}{24}}\implies \csc x=\sqrt{\frac{8}{3}}\)
\(\sin x=\sqrt{\frac{3}{8}}\)
OK, so you did an arcsine on that?

Answer 13

yes

Answer 14

I did all this..i cant get the solution

Answer 15

I got about \(\pm 38^{\circ}\)

Answer 16

the solution is supposed to be something like pi/3+2kpi

Answer 17

@Luis_Rivera

Answer 18

@jim_thompson5910

Answer 19

Oops. DOH! Factored out wrong!
\(\csc x=\sqrt{\frac{32}{24}}\implies \csc x=\sqrt{\frac{4}{3}}\)
\(\sin x=\sqrt{\frac{3}{4}}\implies \sin x=\frac{\sqrt{3}}{\sqrt{4}}\implies \sin x=\frac{\sqrt{3}}{2}\)
That makes a HUGE difference!

Answer 20

Was trying to replay and it reloaded on me... took forever! Well, close enough to forever.

Answer 21

reply... not replay.

Answer 22

loll..wait so thats it?

Answer 23

There is still the \(\pm\) issue. Because the original was squared we do not know if we are getting \(\frac{\sqrt{3}}{2}\) or \(-\frac{\sqrt{3}}{2}\) at that point. So on the unit circle, there are two possible answers.

Answer 24

So, do you know what angles those will be when solved for?

Answer 25

60 and -60?

Answer 26

well it says that it has to be between 0 and 2pi so it has to be the postive one

Answer 27

Yes, which are \(\pm\frac{\pi}{3}\). Hmmm. So it is either \(\frac{\pi}{3}\) or that and the positive coterminal of \(-\frac{\pi}{3}\) which is \(\frac{5\pi}{3}\)

Answer 28

so how would the final solution look like with the correct form?

Answer 29

Is it solve for every angle, and just use \([0,2\pi)\) as the basis?

Answer 30

Or is it just find solutions between 0 and 2pi? Because that +k stuff is only on the first one.

Answer 31

no you have to find all the solutions...if you dont know I can ask someone else

Answer 32

@.Sam. help?

Answer 33

Oh, no. I know. hehe. Just clarifying. If it is all answers, then we need the \(\frac{5\pi}{3}\) one as well.

Answer 34

wow this took 42 min lollll

Answer 35

\(x=\left\{\left(\frac{\pi}{3}+2k\pi\right),\left(\frac{5\pi}{3}+2k\pi\right)|k\in\mathbb{R}\right\}\)

Answer 36

Well, my making a mistake in the middle and the site reloading on me 3 times did not help how long it took. LOL

Answer 37

i did that..and it says that it is incorrect

Answer 38

Do they use union in this one?

Answer 39

no just list them

Answer 40

from smallest to largest value

Answer 41

Then technically, it should be \((\frac{\pi}{3}+2k\pi),(\frac{5\pi}{3}+2k\pi)\)... hmmm... possibly without the ( ). Or they did not account for the \(\pm\) and went with only \(\frac{\pi}{3}+2k\pi\). yah, that is smal;lers to largest...

Wait! That usually means all in the range, not all in the universe! Try just \(\frac{\pi}{3},\frac{5\pi}{3}\)

Answer 42

hmmm...
http://www.wolframalpha.com/input/?i=24%28csc%28x%29%29^2%3D32&dataset=

Answer 43

pi/3 worked but 5p/3 did not

Answer 44

They are showing a cycle on pi, rather than 2pi... but the end result would be similar to what I was showing.

OK, so.... pi/3... that is very strange. If you look at the wolfram page, that is different than what the answr should be. LOL