a) Show that $x=2\sqrt{2}\cos(A)$ satisfies the cubic equation $x^3 - 6x = -2$ provided that $\cos(3A)$ = $\frac{-1}{2\sqrt{2}}$
I did not have a difficulty with this question, I have provided it for requisite context.
b) Using the previous result, find the three roots of the equation $$x^3-6x+2=0$$ giving your answer correct to four decimal places.
I had an issue with this question and would love an explanation as to how to solve it. The answer being kept to four decimal places is also trivial.
Take $A=\frac13\arccos\left(-\frac1{2\sqrt2}\right)$. Then $\cos(3A)=-\frac1{2\sqrt2}$, and therefore $2\sqrt2\cos(A)$ is a solution of the given equation. But, by the same reason, so are the numbers $2\sqrt2\cos\left(A+\frac{2\pi}3\right)$ and $2\sqrt2\cos\left(A+\frac{4\pi}3\right)$. So, the three roots that you are after are $2\sqrt2\cos(A)$, $2\sqrt2\cos\left(A+\frac{2\pi}3\right)$ and $2\sqrt2\cos\left(A+\frac{4\pi}3\right)$.