Can anyone help me with some maths?

[Royality]

Sorry to clutter up the forums with more of my mathematical ineptitude but can anyone explain to me how to do the following equation, I've been stuck on it for an hour.

Solve, for 0 < x < 90, the equation 3sin6xcosec2x = 4

I just can't seem to rearrange it to get a meaningful answer.

I got 3sin6x/sin2x but I don't know how to cancel sin2x from sin6x. I tried implementing sin(3x+3x) getting 18sinx - 24sinx^2/2sinxcosx but that cancels to give me 9/cosx - 12tanx = 4 which is equally unhelpful.

I've got a c3 exam in a day and I'm rubbish and I need 80% at least to get an A. I think I may go and cry.

Please help me. Thanks.

 

[Duff-Man]

Hint: cos(x) = sqrt(1 - sin^2(x) )

And you already did the hard bit

...if you need the rest scroll down





























rearrange what you had: 9/cos(x) - 12tan(x) = 4

into: 9 - 12sin(x) = 4cos(x)

so, squaring both sides: [ 9 - 12sin(x) ]^2 = 4[ 1 - sin^2(x) ]

then expand, and solve as a quadratic.

 

[Royality]

rearrange what you had: 9/cos(x) - 12tan(x) = 4

into: 9 - 12sin(x) = 4cos(x)

so, squaring both sides: [ 9 - 12sin(x) ]^2 = 4[ 1 - sin^2(x) ]

then expand, and solve as a quadratic.

Hmmm, well I did actually get that far but I didn't think you could just square sin and cos like that. Thanks for the help!

I do have another question that has been bugging me, but it is literally impossible.

Read and weep:

The region R is bounded by the curve y=2ln(x-1) and the lines x=0, y=0 and y=p. The region R is rotated completely about the y-axis to form a solid.

Show that the volume, V, is given by:

V= pi(e^p + 4e^0.5p + p - 5)

So I realise you have to rearrange the equation y=2ln(x-1) to get x in terms of y. This leaves you with 1 +e^y/e^2 = x (I think!).
Squaring it gives: 1+e^2y/e^4

And that's all I can do. Help would be much appreciated.

 

[Psiko]

You've got a couple of things wrong there. Firstly your re-arranging isn't correct. You should get x = e^(y/2) + 1, and then your squaring of (what you got for) x forgets about the cross-terms - you've just squared each individual term.

Do you know the formula for a volume of revolution? You are essentially just summing the areas of the circles formed when a single point of the function is rotated about the axis, so it's just the integral of the formula for the area of a circle. So V = Int[pi x^2, 0<=x<=p]dy. Now put x in terms of y and integrate.

 

[Royality]

Can anyone help me with the following integration? To be done by parts:

Integrate - xln(x^2 + 1)

So you can't integrate a ln function so:

u = ln (x^2 + 1) du/dx = 2x\(x^2 + 1)
dv/dx = x v = 1/2x^2

using uv - [v.du/dx] = 1/2x^2.ln(x^2 + 1) - 1/6x^3.ln|x^2 + 1|

Is that right? I think I've made some integration errors. Can someone help me at all? Thanks guys. xxx