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    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
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    (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.

    Here's the relevant links.

    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
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    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
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    (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 t^3-3t^2-3t+1=0 to get t^3-3t^2=3t+1.

    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 (t^3-3t^2)^2-(3t+1)^2=0.

    Expanding this and substituting t^2=s^2-1 will give you a polynomial of degree 6.

    Substituting x=s^2 will give you what you're after.

    Darren
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    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
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    (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
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    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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