<s>[INST] Finding roots of cubic (trig) 

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?
 [/INST] 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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