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.

Thee trigonometric method relies on the trigonometric identity:
$$\cos 3\theta=4\cos^3\theta -3\cos\theta.$$
Set $\;y=A\cos\theta,\enspace A>0$. The equation becomes
$$A^3\cos^3\theta -3A\cos\theta=\sqrt2$$
We choose $A>0$ such that the left-hand side of the equation is proportional to the development of $\cos3\theta $, i.e.
$$\frac{A^3}4=\frac{3A}3\iff A^2=4.$$
Thus we set $y=2\cos\theta $, so that we get the trigonometric equation
$$2\cos3\theta =\sqrt2\iff \cos3\theta =\frac{\sqrt2}2=\cos\frac\pi4,$$
and the solutions are
$$3\theta \equiv\pm\frac\pi4\mod2\pi\iff\theta \equiv\pm\frac\pi{12}\mod\frac{2\pi}3$$
There are $3$ different values for $\cos\theta$, and the different values for $y$ are
$$2\cos\frac\pi{12},\quad2\cos\frac{3\pi}4=-\sqrt2,\quad2\cos\frac{-7\pi}{12}=-\sin\frac\pi{12}$$
Now, the addition formulae yield:
\begin{align*}
\cos\frac\pi{12}&=\cos\Bigl(\frac\pi3-\frac\pi4\Bigr)=\dots=\frac{\sqrt6+\sqrt2}4,\\
\sin\frac\pi{12}&=\sin\Bigl(\frac\pi3-\frac\pi4\Bigr)=\dots=\frac{\sqrt6-\sqrt2}4.
\end{align*}
