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The question is

By putting $x$ $=$ $\frac 23 \cos (\theta)$ Find the exact roots of the equation in terms of $\pi$

$$ 27x^3 - 9x = 1 $$


What I have attempted:

$$ 27x^3 - 9x = 1 $$

$$x=\frac 23 \cos (\theta)$$

$$ x^3=\frac 8{27} \cos^3 (\theta)$$

$$ \therefore 27\left(\frac 8{27} \cos^3 (\theta)\right) - 9\left(\frac 23 \cos (\theta)\right) = 1$$

$$ 8\cos^3(\theta) - 6\cos(\theta) - 1 = 0 $$

Now I tried letting $\cos(\theta)$ = $z$ and try solving the cubic but the solutions aren't exactly nice looking (rational). Is there a special trig identity that I can reduce this equation to?

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Recall that

$$\cos^3 \theta = \frac 3 4 \cos \theta + \frac 1 4 \cos 3\theta$$

Hence you can rewrite your equation as

\begin{align*} 0 &= 8 \cos^3 \theta - 6 \cos \theta - 1 \\ &= 6 \cos \theta + 2 \cos 3\theta - 6 \cos \theta - 1 \\ &= 2 \cos 3\theta + 1 \end{align*}

and hence reduce to $\cos 3\theta = -\frac 1 2$. After a little bit more work, you can write the root of the original polynomial as $\cos r\pi$ with $r$ a reasonably nice fraction.

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  • $\begingroup$ You seem to have accidentally flipped a sign. $\endgroup$ – Matt Samuel Jan 24 '16 at 1:38
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A start: Rewrite as $4\cos^3 \theta-3\cos\theta=\frac{1}{2}$ and recall that $\cos(3\theta)=4\cos^3\theta-3\cos\theta$.

Now we know $\cos(3\theta)$, and can find the three possibilities for $\cos\theta$.

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