<s>[INST] Solve $y^3-3y-\sqrt{2}=0$ using trigonometry This is a part of a larger question.
I had to show that for $4x^3-3x-\cos 3\alpha=0$ one of the solutions is $\cos \alpha$ and then find the other two solutions. Here they are:
$$4x^3-3x-\cos 3\alpha = (x-\cos \alpha)(2x+\cos \alpha + \sqrt{3} \sin \alpha)(2x+\cos \alpha - \sqrt{3} \sin \alpha)$$
I have to use the above and the following results: $\cos 15^{\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2\sqrt{2}}$ to find the solutions of the following:
$$y^3-3y-\sqrt{2}=0$$
I assumed that the constant term must be the equivalent of the cosine term and tried to find alpha so that I have one solution and then can derive the other. But since $\arccos {\sqrt{2}}$ is not trivially defined, this is not a correct approach. Or at least it is not correct the way I am doing it. Also I would have to do a bit more trigs since the second polynomial is not equivalent to the first one. There must be an easier, neater solution.
 [/INST] Recall the triplication formula:
$$
\cos3\alpha=4\cos^3\alpha-3\cos\alpha
$$
so, if your equation is $y^3-3y=\sqrt{2}$, you can first set $y=at$, so
$$
a^3t^3-3at=\sqrt{2}
$$
and you'd like that $a^3/3a=4/3$, so you can take $a=2$: $8t^3-6t=\sqrt{2}$; setting $t=\cos\alpha$, we get
$$
\cos3\alpha=\frac{\sqrt{2}}{2}
$$
so
$$
3\alpha=\frac{\pi}{4}+2k\pi
$$
where $k=0$, $k=1$ or $k=2$. Hence
$$
\frac{\pi}{12},\quad\frac{3\pi}{4},\quad\frac{17\pi}{12}
$$
are the solutions for $\alpha$.
Now,
$$
2\cos\frac{\pi}{12}=2\sqrt{\frac{1+\cos(\pi/6)}{2}}=
\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}
$$
then
$$
2\cos\frac{3\pi}{4}=-\sqrt{2}
$$
and
$$
2\cos\frac{17\pi}{12}=-2\cos\frac{5\pi}{12}=
-2\sqrt{\frac{1+\cos(5\pi/6)}{2}}=
-\sqrt{2-\sqrt{3}}=\frac{\sqrt{6}-\sqrt{2}}{2}
$$
</s>