How would you find the trigonometric roots of a cubic? 
Question: How would you find the roots of the cubic$$x^3+x^2-10x-8=0\tag{1}$$


I'm not too sure where to begin. I'm thinking of somehow, implementing $\cos 3\theta=4\cos^3\theta-3\cos\theta$.
I've tried substituting $x$ with $t+t^{-1}$, but didn't get anywhere, and using Vieta's trigonometric solution formula, I got 
$$x_1=\frac {2\sqrt{31}}3\cdot\cos\left(\frac {\arccos \frac {2}{\sqrt{31}}}3\right)-\frac 13\\x_2=\frac {2\sqrt{31}}3\cdot\cos\left(\frac {2\pi+\arccos\frac {2}{\sqrt{31}}}3\right)-\frac 13\\x_3=\frac {2\sqrt{31}}{3}\cdot\cos\left(\frac {4\pi+\arccos\frac 2{\sqrt{31}}}3\right)-\frac 13$$
But that's not the form I want. I'm looking for a form of $2\left(\cos\frac {\text{something}}{31}+\cos\frac {\text{something}}{31}+\cos\frac {\text{something}}{31}\right)$.
I've spent so much time, that I'm practically burnt out. -.-

1) Reduce the standard way your equation $x^3+x^2-10x-8=0$ to the form $$x^3+ax+b=0\qquad(*)$$
2) Since you have the identity $$4\cos^3\theta-3\cos \theta-\cos 3\theta=0\qquad(**)$$ make $x=u\cos\theta$ in $(*)$ so you get
$$u^3\cos^3\theta+au\cos\theta+b=0\iff4\cos^3\theta+\frac{4au}{u^3}\cos\theta+\frac{4b}{u^3}\qquad(***)$$
3) In order to refer $(***)$ to $(**)$ you need to take $$-3=\frac{4au}{u^3}\iff u=2\sqrt{\frac{-2a}{3}}$$ and $$\frac{4b}{u^3}=-\frac{3b}{2a}\sqrt{\frac{-3}{a}}$$ which gives $$\cos3\theta=\frac{3b}{2a}\sqrt{\frac{-3}{a}}$$
Hence your roots are given by $$x_k=2\sqrt{\frac{-a}{3}}\cos\frac{\theta_k}{3}$$ where $$\theta_k=\arccos\left(\frac{3b}{2a}\sqrt{\frac{-3}{a}}-\frac{2k\pi}{3}\right);\space k=0,1,2$$
Note.-You can use this method because the three roots of your equation are real.
