I was trying to solve a cubic equation using trigonometric representation of Cardano's Formula solutions.
My equation looks like this:
$$x^3-3mx+6m=0$$
I can only find $\cos(\theta)$ however to find the solutions I need $\cos \left(\frac {\theta}{3}\right)$. This can be done easily if I had numeral values for the coefficients however this is not the case here. All I know is that $m$ is a positive constant.
How can I proceed further to find it and also $\cos \left(\frac {\theta+4 \pi}{3}\right)$ and so on...
Let $x= 2\sqrt m \cos\theta$ and substitute into $x^3-3mx+6m=0$ to get
$$4\cos^3\theta - 3\cos\theta =-\frac3{\sqrt{m}} $$ which, per the identity $4\cos^3\theta - 3\cos\theta =\cos3\theta$, leads to $\cos3\theta = - \frac3{\sqrt{m}}$, or $\theta = \frac13 \cos^{-1}( - \frac3{\sqrt{m}} )+\frac{2\pi k}3$. Thus, the solutions are
$$x=2\sqrt m \cos \left(\frac13 \cos^{-1}( - \frac3{\sqrt{m}} )+\frac{2\pi k}3 \right), \>\>\>k=0,1,2$$