Maths! Need Urgent Help!

[Applecrusher]

HELP i have to hand in some work tommorow and have a few q's that i need help with!

1) Solve for x across [0,2pi]
5cos(2x-(pi/6))=2


2)Choose the correct answer. if f:R\{1} -> where f(x) = (2x+3)/(x-1) then f-1 (x) is:
(A) (x+3)/(x-1), xE R\{1} (B) 5/(x-2)+1, xE R\{2}
(C) 5/(x-2)+1, xE R\{1} (D) 5/(x-1)+1, xE R\{1}
(E) (x+3)/(x-1), xE R

Any help on how todo these would be greaty appreciated!
ALSO the answer to the second one is B i just dont know how to get it..

[Applecrusher]

Anyone?

[Auran]

I'm assuming that you know to integrate the equation in question 1 across the range 0 to 2pi. If you look on the internet there are a lot of tutorials on this sort of thing that will go through step by step on how to integrate different functions. Unfortunately I can't remember off the top of my head what cos integrates to.

As for the second one, would have to look back at my maths notes for that one and unfortunately I'm at work and they're 60miles away stashed in my parents loft So can't help you there.

[Big RICHARD]

It's not integration. It's looking for the values of x which satisfy the equation between the limits......

5cos(2x-pi/6)=2

cos(2x-pi/6)=5/2

2x-pi/6=cos^-1(0.4)

2x=pi/6 + cos^-1(0.4)

x=(pi/6 + cos^-1(0.4))/2

Which on the graph of y=cos x has 2 solutions between 0 and 2pi

[loki]

for the first one,
(0, 2pi) is eqaul to (0, 360) if i remember corectly

and the equasion

5cos(2x-(pi/6)) = 2
5cos2x - 5cospi/6 = 2, but since pi/6 = 30 degrees, we then have cos 30 = root3/2
5cos2x - 5root3/2 = 2,
5cos2x = 2 + 5root3/2
cos2x = 2/5 + 5root3/10
cos2x = 2/5 + root3/2
cosx = 2/10 + root3/4
cosx = 0.633 (to 3 decimal places)
x = cos-1(0.633)
x = 50.7 (to 3 significant places)

but since it was asked between 0, 360
x will be equal to 50.7 and 309.3 since x is positive and cos is positive in the 1st and 4th quardants.

I wouldnt take this answer as gospel but i do believe it is correct (if pi is indeed equal to 180 degrees, i cant remember perfectly, been like 5 years since i did this)

[Big RICHARD]

No, loki, because cos(x+pi/6) does not equal cos(x) + cos(pi/6)

[Big RICHARD]

I don't recognise the notation for question 2. If you tell me what it means I'll get the solution....

[loki]

you'll be right, i havent done math since i was 15/16 and i'm 21 now, so i dont imagine my answer will be correct.

at least i remembered i needed 2 answers and i was going about it the right way,shame i forgot that you cant seperate brackets with trigonometric functions.

[Auran]

Oops my mistake, had I read it a bit more carefully, I'd have spotted what it was really after. No wonder my maths teachers always dispaired of me

[Applecrusher]

ummm the f:R\{1} -> R bit in the second q means... function such that real excluding 1 maped on to a real range.... me thinks...
f^-1(x) is the inverse function of f(x)
and the E in the answers is an 'element' of....

[paradidle]

Yes and x to the power y! WORD!! yo!

[fraggie]

if u still need the first one doin u do...

5cos (2x-(pi/6) = 2
cos (2x-pi/6) = 2/5
2x-(pi/6) = (cos^-1) 2/5
and becus ure getting 2x - pi/6 u need to double the top amount of the limit and take off pi/6 to it :. its x = (cos^-1) 2/5 between 0 and 23/6pi
therefore to get x u need to divide the amounts u get between these limits by 2 and then add the pi/6 and u shud get the values of x



i think

[TiG]

Just 1 thing to bear in mind from memory shouldn't all the manipulation be done in Radians with the fomulas, there being 2Pi Radians in 360 degrees. The thing that said that to me is the pi/6 element.

TiG

[Applecrusher]

YAY
i have got the first one (thanx to y'all) and the second one from a friend

THANX ALL!

[fraggie]

how did ya do the 1st one? prob not how i did it lol