What values for,(posted below) satisfy the equation?

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What values for,(posted below) satisfy the equation?

Mathematics
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I don't understand what the question needs besides from 0< x< 2pi means from 0 to 360 degrees which is the full circle
there is a trig identity \[\tan^2(\theta) = 1-\sec^2(\theta) \]
What values for, |dw:1436466383880:dw| satisfy the equation?
maybe we can replace that tan with what I've wrote... ??? dah I've never seen this formatted question before x.x |dw:1436502474870:dw|
The rest of the equation is in the second attachment at the top
|dw:1436466531453:dw|
this is what I'm thinking but I'm not 100% sure ... use the identity on the left hand side \[\tan^2(\theta) = 1-\sec^2(\theta) \]
and then add \[\frac{3}{2}\sec(\theta) \] to both sides so we end up with some sort of quadratic equation (well similar )
ehhh im not so sure how to do this one either.
|dw:1436466790578:dw|
Those are my answer options
HI do what was suggested above write it as a quadratic equation in terms of secant then solve
I don't know how to do these equations at all
lol then cheat!
haha well i was thinking you guys could give me the right answer cuz i cant find it
i 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%3D-3%2F2sec%28x%29
yes @misty1212 we write it as a quadratic equation in terms of secant and then solve
@misty1212 anything can help, i just need the right answer
I know most of the process... but towards the end . I might crash
I think it has something to do with adding 2pi or subtracting 2pi ...hence more than one answer
|dw:1436503293700:dw|
|dw:1436503337897:dw|
|dw:1436503367107:dw| now multiply the entire equation by -2 to get rid of the fraction and the negative
idk, i dont have a calculator at the moment sadly lol
|dw:1436503536319:dw|
|dw:1436503556177:dw|
now let all secants be x for now because that quadratic equation looks mean
|dw:1436503613776:dw|
oh yes we can factor this woo!
ok im learning lol
|dw:1436503685977:dw|
I'm using b^2-4ac so I would know what that part is already for that quadratic formula
|dw:1436503730219:dw|
|dw:1436503768417:dw|
|dw:1436503804424:dw|
So how do we find one of my answers from A-D? lol
\[(\sec(\theta)-2)(\sec(\theta)+\frac{1}{2})=0\] something is screwed up... I am expanding and the last term isn't working out
nice I found the problem -_-! only one of them is right
but strange taking the quadratic equation approach always works...
Can you explain, or do you just wanna give the answer?
-__- *facepalm* I made that error by not getting rid of that fraction .. argh.. ok I thought we were done for.
\[(\sec(\theta)-2)(2\sec(\theta)+1)=0\]
so now we have to solve this separately . . . which I swear we have already done it when I found my roots.
Alrightyyy
\[\sec(\theta)-2=0\] adding 2 to both sides \[\sec(\theta) =2 \] \[(2\sec(\theta)+1)=0 \] that's \[\sec(\theta) = \frac{-1}{2} \]
turns out that we already did this step and not even know it ^ because I've used the quadratic equation to get its roots.
so secant = 1/cosine this would be a bit easier to work with
ok whats next
so my cosine has to be \[\frac{1}{2} \] for \[\sec(\theta) =2 \] and to check secant of 1/2 = 1/(1/2) = 2
|dw:1436504658208:dw| based on that triangle cosine 1/2 is 60 degrees otherwise known as \[\frac{\pi}{3} \] in radians
So is the answer D?
yeah, but I want to get the other answer from that choice.
we got our pi/3. let's see if I can grab 5pi/3
which is 300 degrees.
OH 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.
\[\frac{3}{3} \times 2\pi-\frac{\pi}{3} = \frac{6 \pi-\pi}{3} \]
\[\frac{5\pi}{3} \]
there. XDDDDDDDDDDDDDDDDDDDDDDD
I can't believe I just did that ... it's been 6 years....
yay you did it lol
the 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.
yeah its tough sometimes, but at least you guys are great help
thanks :D
you're welcome. i gave you a medal

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