The specific problems are trig ones:

Find all of the solutions of the following equations in the interval [0,2π)
(3tan^2)2x-1=0

csc^2 (πx/2)-4=0

tan^x*cscx=tan^2 x

2cos(x/2)-3^(1/3)=0

2sin^2 x+5sinx-3=0

Senty is the man for this. But I *think* I've worked out the first one... I haven't done Maths for two years so hopefully you have an answer book where you can check this? If not someone else correct me.

(3tan^2)2x-1=0
(3tan^2)2x = 1
3(tan2x)^2 = 1
tan2x^2 = 1/3
tan2x = SQR(1/3)
2x = tan^-1 SQR(1/3)
2x = 30
x = 15 (degrees)
x = 0.262 (rad)

You are getting your maths HW done from us? Smart Naughty girl.
I know trig pretty well. And I am pretty sure I can solve all that (because they look simple), but the problem being, I am finding it difficult to make head or tail of this typed stuff. I need to write it properly down and try. I could do it by today evening.

Let me introduce you to the genius that is Wolfram Alpha. This is the final problem you listed:
http://www.wolframalpha.com/input/?i=2* ... +-+3+%3D+0

Your notation is a little inconsistent, so you'll have to check that it reads it correctly.

Incidentally, Gladstone's solution for the first one is correct, but fails to take into account the negative square root and the repetition period of the function. Solutions are thus given by x = (1/12) (6 pi n - pi) and x = (1/12) (6 pi n + pi), for all whole number values of n that give x in the required range (between 0 and 2 pi). Thus, there are solutions at pi/12 (the one he found), 7pi/12, 13pi/12, 19pi/12, 5pi/12, 17pi/12, and 23pi/12.

Ew ew EW.
MATH.

Sentynel wrote:Let me introduce you to the genius that is Wolfram Alpha. This is the final problem you listed:
http://www.wolframalpha.com/input/?i=2* ... +-+3+%3D+0

That one is brilliant.

Sentynel wrote:Let me introduce you to the genius that is Wolfram Alpha. This is the final problem you listed:
http://www.wolframalpha.com/input/?i=2* ... +-+3+%3D+0

Your notation is a little inconsistent, so you'll have to check that it reads it correctly.

Incidentally, Gladstone's solution for the first one is correct, but fails to take into account the negative square root and the repetition period of the function. Solutions are thus given by x = (1/12) (6 pi n - pi) and x = (1/12) (6 pi n + pi), for all whole number values of n that give x in the required range (between 0 and 2 pi). Thus, there are solutions at pi/12 (the one he found), 7pi/12, 13pi/12, 19pi/12, 5pi/12, 17pi/12, and 23pi/12.

Ah. Niceness.

Holycabbage




I love math.

-EDIT-

FASDFURH BEST CENSOR EVER I'VE NEVER LAUGHED THIS HARD

Huh?

flooping 'eck
cabbage
Bullcabbage
cabbage
Damn
Ass
Bitch

---

What censors are you speaking of? I'm confused.

wrote:What censors are you speaking of? I'm confused.

footnoodles wrote:Holycabbage

Take a guess, unless you do actually use the word 'cabbage' in place of ... something else.

Ah. Somehow I skipped over the first paragraph. I thought the censored swear was 'FASDFURH'

That would be a good censor too.