Are there any ways to convert inverse trigonometric values to radicals? When we solve a cubic equation $ax^3+bx^2+cx+d=0$, the roots are supposed to be in the form of radicals in real numbers or complex realm. However,  if the discriminant is less than 0, the solution is ended up with roots represented by inverse trigonometric function in most cases. For example, the three roots for $x^3−4x+1=0$ are all in trigonometric form. And the equation $x^3−2x+1=0$ has 1 rational root, and two other roots that could be in radical form if solved by factorization method or inverse trigonometric values if solved by Cardano's solution and trigonometric method. By comparing their decimals, the roots obtained by two different methods are equal. My question is - are there any general ways to convert these inverse trigonometric values to radicals?
 A: For a cubic equation when the discriminant is less than zero, the roots may be expressed in the form of trigonometric function of an angle in inverse trigonometric form if solved by Cardano method. For example, $x^3−2x+1=0$
\begin{cases} x_1=2\sqrt{\dfrac{2}{3}} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\big(-\dfrac{3}{4}\sqrt{\dfrac{3}{2}}\big)\bigg]   \\  x_2=2\sqrt{\dfrac{2}{3}} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\big(-\dfrac{3}{4}\sqrt{\dfrac{3}{2}}\big)+\dfrac{2\pi}{3}\bigg]  \\ x_3=2\sqrt{\dfrac{2}{3}}  \cos \bigg[ \dfrac{1}{3}\cdot \arccos\big(-\dfrac{3}{4}\sqrt{\dfrac{3}{2}}\big)+\dfrac{4\pi}{3} \bigg]   \end{cases}
However, the roots may be given in a very clean radical form if solved by factorization method.
\begin{cases} x_1 = 1\\x_2 =-\dfrac{1}{2}-\dfrac{\sqrt{5}}{2}  \\   x_3=-\dfrac{1}{2}+\dfrac{\sqrt{5}}{2} \end{cases}
If one of the roots is rational, the trigonometric forms could be converted to radical form.
In this case, given $\cosθ = -\dfrac{3}{4}\sqrt{\dfrac{3}{2}}$
, then  by using the identity $4\cos^3θ-3\cos θ-\cos3θ =0$, $\cos\dfrac{θ}{3}=\dfrac{1}{2}\sqrt{\dfrac{3}{2}}$
Then $x_2 = 2\sqrt{\dfrac{2}{3}} \cos \bigg[ \dfrac{θ}{3} +\dfrac{2\pi}{3}\bigg] =-\dfrac{1}{2}-\dfrac{\sqrt{5}}{2}$
For those cubic equations without rational root, for example, $ x^3−4x+1=0$, the roots look like
\begin{cases} x_1=4\sqrt{\dfrac{1}{3}}\cos \bigg[\dfrac{1}{3}\cdot \arccos\big(-\dfrac{3}{16}\sqrt{3}\big)\bigg]   \\  x_2=4\sqrt{\dfrac{1}{3}} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\big(-\dfrac{3}{16}\sqrt{3}\big)+\dfrac{2\pi}{3}\bigg]  \\ x_3=4\sqrt{\dfrac{1}{3}}  \cos \bigg[ \dfrac{1}{3}\cdot \arccos\big(-\dfrac{3}{16}\sqrt{3}\big)+\dfrac{4\pi}{3} \bigg]   \end{cases}
which may take up broad number of cases, still remains unresolved.
