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Roots and trig

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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 sin(15)=3−122\displaystyle sin(15)=\frac{\sqrt3 -1}{2\sqrt2}

b) I said 4x3−3x=cos(3α) \displaystyle 4x^3-3x=cos(3\alpha)

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

4x3−3x=4cos3(α)−3cos(α)\displaystyle 4x^3-3x=4cos^3(\alpha)-3cos(\alpha) so x=cos(α)\displaystyle x=cos(\alpha) is a root.

Other two roots are; x=−cos(α)±3sin(α)2\displaystyle x=\frac{-cos(\alpha)\pm \sqrt3 sin(\alpha)}{2}.
Original post by RDKGames
aefaefr.PNG

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 sin(15)=3−122\displaystyle sin(15)=\frac{\sqrt3 -1}{2\sqrt2}

b) I said 4x3−3x=cos(3α) \displaystyle 4x^3-3x=cos(3\alpha)

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

4x3−3x=4cos3(α)−3cos(α)\displaystyle 4x^3-3x=4cos^3(\alpha)-3cos(\alpha) so x=cos(α)\displaystyle x=cos(\alpha) is a root.

Other two roots are; x=−cos(α)±3sin(α)2\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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Avatar for Zacken
Zacken
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Original post by RDKGames
...


12y3−3y2−12=0\frac{1}{2}y^3 - \frac{3y}{2} - \frac{1}{\sqrt{2}} = 0

⇒4(12y)3−3(12y)−12=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⁡3α=12\cos 3\alpha = \frac{1}{\sqrt{2}} so the above is...
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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