anonymous
  • anonymous
What values for,(posted below) satisfy the equation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
This fills in the blank
1 Attachment
anonymous
  • anonymous
This follows the question
1 Attachment
anonymous
  • anonymous
@UsukiDoll

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More answers

UsukiDoll
  • UsukiDoll
I don't understand what the question needs besides from 0< x< 2pi means from 0 to 360 degrees which is the full circle
UsukiDoll
  • UsukiDoll
there is a trig identity \[\tan^2(\theta) = 1-\sec^2(\theta) \]
anonymous
  • anonymous
What values for, |dw:1436466383880:dw| satisfy the equation?
UsukiDoll
  • UsukiDoll
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|
anonymous
  • anonymous
The rest of the equation is in the second attachment at the top
anonymous
  • anonymous
|dw:1436466531453:dw|
UsukiDoll
  • UsukiDoll
this is what I'm thinking but I'm not 100% sure ... use the identity on the left hand side \[\tan^2(\theta) = 1-\sec^2(\theta) \]
UsukiDoll
  • UsukiDoll
and then add \[\frac{3}{2}\sec(\theta) \] to both sides so we end up with some sort of quadratic equation (well similar )
oleg3321
  • oleg3321
ehhh im not so sure how to do this one either.
anonymous
  • anonymous
|dw:1436466790578:dw|
anonymous
  • anonymous
Those are my answer options
misty1212
  • misty1212
HI do what was suggested above write it as a quadratic equation in terms of secant then solve
anonymous
  • anonymous
I don't know how to do these equations at all
misty1212
  • misty1212
lol then cheat!
anonymous
  • anonymous
haha well i was thinking you guys could give me the right answer cuz i cant find it
misty1212
  • misty1212
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
UsukiDoll
  • UsukiDoll
yes @misty1212 we write it as a quadratic equation in terms of secant and then solve
anonymous
  • anonymous
@misty1212 anything can help, i just need the right answer
UsukiDoll
  • UsukiDoll
I know most of the process... but towards the end . I might crash
UsukiDoll
  • UsukiDoll
I think it has something to do with adding 2pi or subtracting 2pi ...hence more than one answer
UsukiDoll
  • UsukiDoll
|dw:1436503293700:dw|
UsukiDoll
  • UsukiDoll
|dw:1436503337897:dw|
UsukiDoll
  • UsukiDoll
|dw:1436503367107:dw| now multiply the entire equation by -2 to get rid of the fraction and the negative
anonymous
  • anonymous
idk, i dont have a calculator at the moment sadly lol
UsukiDoll
  • UsukiDoll
|dw:1436503536319:dw|
UsukiDoll
  • UsukiDoll
|dw:1436503556177:dw|
UsukiDoll
  • UsukiDoll
now let all secants be x for now because that quadratic equation looks mean
UsukiDoll
  • UsukiDoll
|dw:1436503613776:dw|
UsukiDoll
  • UsukiDoll
oh yes we can factor this woo!
anonymous
  • anonymous
ok im learning lol
UsukiDoll
  • UsukiDoll
|dw:1436503685977:dw|
UsukiDoll
  • UsukiDoll
I'm using b^2-4ac so I would know what that part is already for that quadratic formula
UsukiDoll
  • UsukiDoll
|dw:1436503730219:dw|
UsukiDoll
  • UsukiDoll
|dw:1436503768417:dw|
UsukiDoll
  • UsukiDoll
|dw:1436503804424:dw|
anonymous
  • anonymous
So how do we find one of my answers from A-D? lol
UsukiDoll
  • UsukiDoll
\[(\sec(\theta)-2)(\sec(\theta)+\frac{1}{2})=0\] something is screwed up... I am expanding and the last term isn't working out
anonymous
  • anonymous
@satellite73 sos
UsukiDoll
  • UsukiDoll
nice I found the problem -_-! only one of them is right
UsukiDoll
  • UsukiDoll
but strange taking the quadratic equation approach always works...
anonymous
  • anonymous
Can you explain, or do you just wanna give the answer?
UsukiDoll
  • UsukiDoll
-__- *facepalm* I made that error by not getting rid of that fraction .. argh.. ok I thought we were done for.
UsukiDoll
  • UsukiDoll
\[(\sec(\theta)-2)(2\sec(\theta)+1)=0\]
UsukiDoll
  • UsukiDoll
so now we have to solve this separately . . . which I swear we have already done it when I found my roots.
anonymous
  • anonymous
Alrightyyy
UsukiDoll
  • UsukiDoll
\[\sec(\theta)-2=0\] adding 2 to both sides \[\sec(\theta) =2 \] \[(2\sec(\theta)+1)=0 \] that's \[\sec(\theta) = \frac{-1}{2} \]
UsukiDoll
  • UsukiDoll
turns out that we already did this step and not even know it ^ because I've used the quadratic equation to get its roots.
UsukiDoll
  • UsukiDoll
so secant = 1/cosine this would be a bit easier to work with
anonymous
  • anonymous
ok whats next
UsukiDoll
  • UsukiDoll
so my cosine has to be \[\frac{1}{2} \] for \[\sec(\theta) =2 \] and to check secant of 1/2 = 1/(1/2) = 2
UsukiDoll
  • UsukiDoll
|dw:1436504658208:dw| based on that triangle cosine 1/2 is 60 degrees otherwise known as \[\frac{\pi}{3} \] in radians
anonymous
  • anonymous
So is the answer D?
UsukiDoll
  • UsukiDoll
yeah, but I want to get the other answer from that choice.
UsukiDoll
  • UsukiDoll
we got our pi/3. let's see if I can grab 5pi/3
UsukiDoll
  • UsukiDoll
which is 300 degrees.
UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
\[\frac{3}{3} \times 2\pi-\frac{\pi}{3} = \frac{6 \pi-\pi}{3} \]
UsukiDoll
  • UsukiDoll
\[\frac{5\pi}{3} \]
UsukiDoll
  • UsukiDoll
there. XDDDDDDDDDDDDDDDDDDDDDDD
UsukiDoll
  • UsukiDoll
I can't believe I just did that ... it's been 6 years....
anonymous
  • anonymous
yay you did it lol
UsukiDoll
  • UsukiDoll
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.
anonymous
  • anonymous
yeah its tough sometimes, but at least you guys are great help
UsukiDoll
  • UsukiDoll
thanks :D
anonymous
  • anonymous
you're welcome. i gave you a medal

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