<s>[INST] Finding the roots of any cubic with trigonometric roots. 
Question: How would you find the roots of a cubic polynomial whose roots are only expressive in trigonometric forms?


I'm really confused on how you would solve it. Some examples:$$x^3+x^2-2x-1=0\\x_1=2\cos\frac {2\pi}7,x_2=2\cos\frac {4\pi}7,x_3=2\cos\frac {8\pi}7\tag{1}$$$$x^3-x^2-9x+1=0\\x_1=4\cos\frac {2\pi}7,x_2=4\cos\frac {4\pi}7+1,x_3=4\cos\frac {8\pi}7+1\tag{2}$$$$x^3+x^2-6x-7=0\\x_1=2\left(\cos\frac {4\pi}{19}+\cos\frac {6\pi}{19}+\cos\frac {10\pi}{19}\right)\\x_2=2\left(\cos\frac {2\pi}{19}+\cos\frac {14\pi}{19}+\cos\frac {16\pi}{19}\right)\\x_3=2\left(\cos\frac {8\pi}{19}+\cos\frac {12\pi}{19}+\cos\frac {20\pi}{19}\right)\tag{3}$$
With $(3)$ being a very famous relation with this problem.
Now that I know it's possible, I'm wondering if there is a simple way to find the roots of any cubic with trigonometric roots. And is it possible to use the method to find the roots of cubics such as $x^3+x^2-10x-8=0$?
 [/INST] The strategy to solve a cubic equation is first to change it to the form $x^3+px+q=0$, i.e. to eliminate the squared term.
Cardano's method works well if there is only $1$ real root ($4px^3+27q^2>0$). If there are $3$ real roots ($4px^3+27q^2<0$), you can set $x=A\cos\theta$ ($A>0$). The equation becomes
$$A^3\cos^3\theta+pA\cos\theta+q=0.$$
We choose $A$ so that $A^3\cos^3\theta+pA\cos\theta$ is proportional to the expansion of $\cos 3\theta$ in function of $\cos\theta$. Remember $\cos 3\theta=4\cos^3\theta-3\cos\theta$. So we must have
$$\frac{A^3}4=\frac{pA}{-3}\iff A^2=\frac{-p}3\iff A=\sqrt{\frac{-p}3}\qquad\text{(we chose $A>0$)}.$$
We obtain
$$-\frac p3\sqrt{-\frac p3}(4\cos^3\theta-3\cos\theta)=-q\iff\cos3\theta=-\frac q{-\dfrac p3\sqrt{-\dfrac p3}}$$
There remains to solve this standard trigonometric equation.
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