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

- anonymous

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

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- anonymous

This fills in the blank

##### 1 Attachment

- anonymous

This follows the question

##### 1 Attachment

- anonymous

@UsukiDoll

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## More answers

- 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

there is a trig identity
\[\tan^2(\theta) = 1-\sec^2(\theta) \]

- anonymous

What values for, |dw:1436466383880:dw| satisfy the equation?

- 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

The rest of the equation is in the second attachment at the top

- anonymous

|dw:1436466531453:dw|

- 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

and then add \[\frac{3}{2}\sec(\theta) \] to both sides so we end up with some sort of quadratic equation (well similar )

- oleg3321

ehhh im not so sure how to do this one either.

- anonymous

|dw:1436466790578:dw|

- anonymous

Those are my answer options

- misty1212

HI
do what was suggested above
write it as a quadratic equation in terms of secant
then solve

- anonymous

I don't know how to do these equations at all

- misty1212

lol then cheat!

- anonymous

haha well i was thinking you guys could give me the right answer cuz i cant find it

- 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

yes @misty1212 we write it as a quadratic equation in terms of secant and then solve

- anonymous

@misty1212 anything can help, i just need the right answer

- UsukiDoll

I know most of the process... but towards the end . I might crash

- UsukiDoll

I think it has something to do with adding 2pi or subtracting 2pi ...hence more than one answer

- UsukiDoll

|dw:1436503293700:dw|

- UsukiDoll

|dw:1436503337897:dw|

- UsukiDoll

|dw:1436503367107:dw| now multiply the entire equation by -2 to get rid of the fraction and the negative

- anonymous

idk, i dont have a calculator at the moment sadly lol

- UsukiDoll

|dw:1436503536319:dw|

- UsukiDoll

|dw:1436503556177:dw|

- UsukiDoll

now let all secants be x for now because that quadratic equation looks mean

- UsukiDoll

|dw:1436503613776:dw|

- UsukiDoll

oh yes we can factor this woo!

- anonymous

ok im learning lol

- UsukiDoll

|dw:1436503685977:dw|

- UsukiDoll

I'm using b^2-4ac so I would know what that part is already for that quadratic formula

- UsukiDoll

|dw:1436503730219:dw|

- UsukiDoll

|dw:1436503768417:dw|

- UsukiDoll

|dw:1436503804424:dw|

- anonymous

So how do we find one of my answers from A-D? lol

- 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

@satellite73 sos

- UsukiDoll

nice I found the problem -_-! only one of them is right

- UsukiDoll

but strange taking the quadratic equation approach always works...

- anonymous

Can you explain, or do you just wanna give the answer?

- UsukiDoll

-__- *facepalm* I made that error by not getting rid of that fraction .. argh.. ok I thought we were done for.

- UsukiDoll

\[(\sec(\theta)-2)(2\sec(\theta)+1)=0\]

- UsukiDoll

so now we have to solve this separately . . . which I swear we have already done it when I found my roots.

- anonymous

Alrightyyy

- 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

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

so secant = 1/cosine this would be a bit easier to work with

- anonymous

ok whats next

- 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

|dw:1436504658208:dw| based on that triangle cosine 1/2 is 60 degrees otherwise known as \[\frac{\pi}{3} \] in radians

- anonymous

So is the answer D?

- UsukiDoll

yeah, but I want to get the other answer from that choice.

- UsukiDoll

we got our pi/3. let's see if I can grab 5pi/3

- UsukiDoll

which is 300 degrees.

- 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

\[\frac{3}{3} \times 2\pi-\frac{\pi}{3} = \frac{6 \pi-\pi}{3} \]

- UsukiDoll

\[\frac{5\pi}{3} \]

- UsukiDoll

there. XDDDDDDDDDDDDDDDDDDDDDDD

- UsukiDoll

I can't believe I just did that ... it's been 6 years....

- anonymous

yay you did it lol

- 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

yeah its tough sometimes, but at least you guys are great help

- UsukiDoll

thanks :D

- anonymous

you're welcome. i gave you a medal

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