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    Not sure how to approach C. Tried saying that cos(3x) is root 2 but that's not valid; so there can't be 3 roots, I know that much from that. Maybe expressing root 2 in terms of cos(15) or sin(15)? But I'm unsure where to proceed from that.

    a) Got \displaystyle sin(15)=\frac{\sqrt3 -1}{2\sqrt2}

    b) I said  \displaystyle 4x^3-3x=cos(3\alpha)

    and after using compound angle formulae on the right side, I got

    \displaystyle 4x^3-3x=4cos^3(\alpha)-3cos(\alpha) so \displaystyle x=cos(\alpha) is a root.

    Other two roots are; \displaystyle x=\frac{-cos(\alpha)\pm \sqrt3 sin(\alpha)}{2}.
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    (Original post by RDKGames)
    Name:  aefaefr.PNG
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    Not sure how to approach C. Tried saying that cos(3x) is root 2 but that's not valid; so there can't be 3 roots, I know that much from that. Maybe expressing root 2 in terms of cos(15) or sin(15)? But I'm unsure where to proceed from that.

    a) Got \displaystyle sin(15)=\frac{\sqrt3 -1}{2\sqrt2}

    b) I said  \displaystyle 4x^3-3x=cos(3\alpha)

    and after using compound angle formulae on the right side, I got

    \displaystyle 4x^3-3x=4cos^3(\alpha)-3cos(\alpha) so \displaystyle x=cos(\alpha) is a root.

    Other two roots are; \displaystyle x=\frac{-cos(\alpha)\pm \sqrt3 sin(\alpha)}{2}.
    start by letting 2x=y and then dividing through by 2
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    (Original post by RDKGames)
    ...
    \frac{1}{2}y^3 - \frac{3y}{2} - \frac{1}{\sqrt{2}} = 0

    \Rightarrow 4\left(\frac{1}{2}y \right)^3 - 3 \left(\frac{1}{2}y \right) - \frac{1}{\sqrt{2}} = 0

    Note that \cos 3\alpha = \frac{1}{\sqrt{2}} so the above is...
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    (Original post by ValerieKR)
    start by letting 2x=y and then dividing through by 2
    Okay that one caught me out completely. Didn't even notice the coefficients weren't the same :/
    Thanks, I'll give this a go now.
 
 
 
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