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anonymous
 one year ago
What values for,(posted below) satisfy the equation?
anonymous
 one year ago
What values for,(posted below) satisfy the equation?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This fills in the blank

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This follows the question

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I don't understand what the question needs besides from 0< x< 2pi means from 0 to 360 degrees which is the full circle

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3there is a trig identity \[\tan^2(\theta) = 1\sec^2(\theta) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What values for, dw:1436466383880:dw satisfy the equation?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3maybe we can replace that tan with what I've wrote... ??? dah I've never seen this formatted question before x.x dw:1436502474870:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The rest of the equation is in the second attachment at the top

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436466531453:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3this is what I'm thinking but I'm not 100% sure ... use the identity on the left hand side \[\tan^2(\theta) = 1\sec^2(\theta) \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3and then add \[\frac{3}{2}\sec(\theta) \] to both sides so we end up with some sort of quadratic equation (well similar )

Oleg3321
 one year ago
Best ResponseYou've already chosen the best response.0ehhh im not so sure how to do this one either.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436466790578:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Those are my answer options

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0HI do what was suggested above write it as a quadratic equation in terms of secant then solve

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know how to do these equations at all

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha well i was thinking you guys could give me the right answer cuz i cant find it

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0i can show you how to find it if you like it will take a bit of time http://www.wolframalpha.com/input/?i=tan%5E2%28x%29%3D3%2F2sec%28x%29

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3yes @misty1212 we write it as a quadratic equation in terms of secant and then solve

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@misty1212 anything can help, i just need the right answer

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I know most of the process... but towards the end . I might crash

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I think it has something to do with adding 2pi or subtracting 2pi ...hence more than one answer

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436503293700:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436503337897:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436503367107:dw now multiply the entire equation by 2 to get rid of the fraction and the negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0idk, i dont have a calculator at the moment sadly lol

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436503536319:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436503556177:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3now let all secants be x for now because that quadratic equation looks mean

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436503613776:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3oh yes we can factor this woo!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436503685977:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I'm using b^24ac so I would know what that part is already for that quadratic formula

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436503730219:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436503768417:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436503804424:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So how do we find one of my answers from AD? lol

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[(\sec(\theta)2)(\sec(\theta)+\frac{1}{2})=0\] something is screwed up... I am expanding and the last term isn't working out

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3nice I found the problem _! only one of them is right

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3but strange taking the quadratic equation approach always works...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you explain, or do you just wanna give the answer?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3__ *facepalm* I made that error by not getting rid of that fraction .. argh.. ok I thought we were done for.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[(\sec(\theta)2)(2\sec(\theta)+1)=0\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so now we have to solve this separately . . . which I swear we have already done it when I found my roots.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[\sec(\theta)2=0\] adding 2 to both sides \[\sec(\theta) =2 \] \[(2\sec(\theta)+1)=0 \] that's \[\sec(\theta) = \frac{1}{2} \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3turns out that we already did this step and not even know it ^ because I've used the quadratic equation to get its roots.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so secant = 1/cosine this would be a bit easier to work with

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so my cosine has to be \[\frac{1}{2} \] for \[\sec(\theta) =2 \] and to check secant of 1/2 = 1/(1/2) = 2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1436504658208:dw based on that triangle cosine 1/2 is 60 degrees otherwise known as \[\frac{\pi}{3} \] in radians

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3yeah, but I want to get the other answer from that choice.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3we got our pi/3. let's see if I can grab 5pi/3

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3which is 300 degrees.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3OH MAN! I am making this hard on myself again ROFL! We just have to do this \[2\pi\frac{\pi}{3} \] because according to our restriction we are in the unit circle 0< theta < 360 means from 0 to 360.. since we need 300 degrees we are in the fourth quadrant.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{3}{3} \times 2\pi\frac{\pi}{3} = \frac{6 \pi\pi}{3} \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3there. XDDDDDDDDDDDDDDDDDDDDDDD

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I can't believe I just did that ... it's been 6 years....

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3the wording on the question threw me off... because in my textbooks it was a bit simpler so I had to think it through for a bit.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah its tough sometimes, but at least you guys are great help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you're welcome. i gave you a medal
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