# A Level maths question help please Watch

1. Hi everyone, can one of you clever people just check the answer to the last part of the question please.

I got X^3 - 6X - 4 = 0

Many thanks
2. (Original post by Owhite01)
Hi everyone, can one of you clever people just check the answer to the last part of the question please.
Something seems off when I attempted verification using WolframAlpha. I haven't personally verified your solution myself though.

http://www.wolframalpha.com/input/?i...83pi%2F4%29%29

http://www.wolframalpha.com/input/?i=x%5E3-6x-4%3D0

I hope this helps.

Darren
3. Hi Darren
Thanks for the help. I forgot to add last night (before I fell asleep) that the cubic equation is solutions for x=t^2. The way I did the question and I'm not sure it's correct but I couldn't think of another way is to use the fact that sec^2x = 1 + tan^2x and thus to put 1+ t^2 for t into their cubic and come up with an equation in t^6, t^4 etc. I then reduced it to a cubic in x by letting t^2 =x. What do you think?
Thanks Paul
4. (Original post by Owhite01)
Hi Darren,

Thanks for the help. I forgot to add last night (before I fell asleep) that the cubic equation is solutions for x=t^2. The way I did the question and I'm not sure it's correct but I couldn't think of another way is to use the fact that sec^2x = 1 + tan^2x and thus to put 1+ t^2 for t into their cubic and come up with an equation in t^6, t^4 etc. I then reduced it to a cubic in x by letting t^2=x. What do you think?

Thanks Paul
Hi Paul,

I fail to see why that would work.

My approach would be to rearrange to get .

Now square both sides of this. You still get the same roots as before as we haven't really done anything.

Taking everything back over to the LHS gives .

Expanding this and substituting will give you a polynomial of degree 6.

Substituting will give you what you're after.

Darren
5. Hi Darren
I've actually realised that my first approach was incorrect. What I have now done is factorise their cubic use the relationship sec^2=1+tan^2 along with the t solutions and multiplied out to get a new equation x^3-18x^2+48x-32. What do you think?
Thanx Paul
6. (Original post by Owhite01)
Hi Darren,

I've actually realised that my first approach was incorrect. What I have now done is factorise their cubic, use the relationship sec^2=1+tan^2 along with the t solutions and multiplied out to get a new equation x^3-18x^2+48x-32. What do you think?

Thanx Paul
Well I'm pretty sure that's the right answer. See above for how I did it.

Darren
7. Hi Darren
Thanks for your solution. I did try a method along your lines but it got too complicated. How would you deal with the term in t^5 when substituting for s^2?
Thanx Paul

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