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Old Dec 15, 2005, 09:02 AM // 09:02   #1
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im paying 5k to anyone who can help me solve this math problem for school lol..
its an algebra 2 class by the way.

3tan2x-sec2x-5=0

the 2's in the equation mean "squared', cus i dont know how to superscript numbers.

anyone who can help solve this or solve this will be paid gw gold. thanks a bunch.
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Old Dec 15, 2005, 09:06 AM // 09:06   #2
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Use a calculator !
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Old Dec 15, 2005, 09:12 AM // 09:12   #3
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Aren't there school teachers for this stuff?
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Old Dec 15, 2005, 09:28 AM // 09:28   #4
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I would help but I made it a point to forget all my algebra after my last math exam 5 years ago.
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Old Dec 15, 2005, 01:32 PM // 13:32   #5
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im in math 4 and id help if i had more time and if i remember what the hell im suppose to do on that problem. if i remember ill get back to u on it
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Old Dec 15, 2005, 02:00 PM // 14:00   #6
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Quote:
Originally Posted by krnkrackmonkey
the 2's in the equation mean "squared', cus i dont know how to superscript numbers.
FYI, using "^2" has the same meaning as a ².
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Old Dec 15, 2005, 02:18 PM // 14:18   #7
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Quick search on Google gives me a rule:
sec²t = 1 + tan²t

So your problem:

sec²x = 5 + 3tan²x

becomes, with substitution;

(1 + tan²x) = 5 + 3tan²x

-4 = 2tan²x

tan²x = -2



And I don't know what the hell tan²x is, I haven't done maths in years...
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Old Dec 15, 2005, 03:12 PM // 15:12   #8
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rolf cheater using google, thats just not even cool, id laugh it was wrong though but it looks rite to me unless u do the opposite of tan and take the square root of the opposite of tan. course i think that would be an imaginary number or i. then u would get x= w/e that answer to that is(dont have a calcualtor handy)
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Old Dec 15, 2005, 03:56 PM // 15:56   #9
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Are you ready for this?

Let's first change everything to sin(x) and cos(x) and move the -5 over to the right hand side. We can do this because tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x) so we have:

3sin²(x)/cos²(x) - 1/cos²(x) = 5

Why am I allowed to do this? Well, tan²(x) is really nothing more than [tan(x)]². That means it's nothing more than [sin(x)/cos(x)]². And now you can bring the squared into the fraction to get sin²(x)/cos²(x). Same thing works for sec²(x)

Now factor out the 1/cos²(x) and you get:

[1/cos²(x)][3sin²(x) - 1] = 5

Then cross multiply to bring the cosine over to the right hand side:

3sin²(x) - 1 = 5cos²(x)

Now, HERE IS WHAT THE TEACHER WANTS YOU TO KNOW. Since sin²(x) + cos²(x) = 1, then sin²(x) = 1 - cos²(x) and we can substitute to get:

3[1 - cos²(x)] - 1 = 5cos²(x)

Now let's distribute and collect like terms:

3 - 3cos²(x) - 1 = 5cos²(x)

and

2 = 8cos²(x)

So...

1/4 = cos²(x)

Now take the square root of both sides. But hold on, there's a catch:

(+/-) 1/2 = cos(x)

That's right. Remember, both -1/2 and 1/2, when squared, will give you 1/4. Sucks, huh?

Guess what that means? There are four goddamn answers to this question. Seriously. Why? Well, there are two values that satisfy cos(x) = 1/2 and two other values that satisfy cos(x) = -1/2, like so:

cos(60) = 1/2
cos(-60) = 1/2
cos(120) = -1/2
cos(240) = -1/2

If your teacher wants the answers in radians, well, I can't do *everything* for you.

Last edited by Raptox; Dec 15, 2005 at 04:28 PM // 16:28..
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Old Dec 15, 2005, 04:00 PM // 16:00   #10
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ummm...wow, youre smart... hahahaha
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Old Dec 15, 2005, 04:01 PM // 16:01   #11
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so ya um rite ok ya good job? um rite ok... i thought u had to square root it since it would be considered incomplete or not in simpliest form
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Old Dec 15, 2005, 04:02 PM // 16:02   #12
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Wow... If your in Math Pure you should have a graphic cal. which will do it for you, just type in the question and it will give you the answer. I hope your not using a $2 cal. for school.
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Old Dec 15, 2005, 04:06 PM // 16:06   #13
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Quote:
Originally Posted by Professor Paladin
ummm...wow, youre smart... hahahaha
If I was smart I'd have figured out how to make the "²" show up without copying it out of Mish's post. -_-
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Old Dec 15, 2005, 06:24 PM // 18:24   #14
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Depending on your current system font, just type alt + 0178 then you will get ².
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Old Dec 15, 2005, 06:49 PM // 18:49   #15
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Math is not my middle name! Gaming is!
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Old Dec 15, 2005, 07:35 PM // 19:35   #16
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°¹²RED ENGINE GO³
¯¯¯¯¯¯
№²°°² ftw!

1,337₧RED ENGINE GORED ENGINE GORED ENGINE GORED ENGINE GO for me!


besides, graphing calculators can solve that for you
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Old Dec 15, 2005, 07:41 PM // 19:41   #17
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o whoa.... math... i almost forgot what that is . i finished university like a year ago, YES UNIVERSITY U COLLEGE BUMBS . lol jokes. any way, math isint in my vocabulary anymore
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Old Dec 15, 2005, 08:32 PM // 20:32   #18
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Either is "isint" heh, joking!
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Old Dec 15, 2005, 08:58 PM // 20:58   #19
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Quote:
Originally Posted by kg_lildude1
°¹²³
¯¯¯¯¯¯
№²°°² ftw!

1,337₧ for me!


besides, graphing calculators can solve that for you
They let us use graphing calculators in high school but gave no credit unless you showed your work. Then in college they didn't let us use calculators at all.

I'd be impressed if the latest TI or HP models gave you all four answers to that problem, though.

Last edited by Raptox; Dec 15, 2005 at 09:01 PM // 21:01..
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Old Dec 15, 2005, 09:13 PM // 21:13   #20
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So I take it this is Pre-cal you are in then? If so then the explination Raptox gave ya is correct. At first I thought you forgot to say you are evaluating limits or taking derivatives.

The best advice I can give ya is to Memorize the Unit Circle!!! If not copy it to the backing of your graphing calculator hehe.

Later on in College most professors dont let you use calculators for the simple fact that you dont really need one to learn the rules of calculus.

Calculus is easy, its always the algebra that can trip you up if not careful.
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