Solving $x^3-3x+\sqrt{3}=0$ (to evaluate $\sin20^\circ$) gives imaginary roots $$ x^3 - 3x + \sqrt{3} = 0 $$
How to solve the above cubic equation?
I used Cardano's Method but, I am getting imaginary roots but, that should not happen because I encountered the above cubic while trying to evaluate $\sin(20^\circ)$.
If not the cubic, please tell me how to evaluate $\sin(20^\circ)$?

The equation 
$$x^3-3x+\sqrt 3=0$$
can be written as
$$-4\left(\frac x2\right)^3+3\cdot\frac x2=\frac{\sqrt 3}{2}$$
Using that
$$\sin(3\theta)=-4\sin^3\theta+3\sin\theta$$
we see that
$$\frac{x}{2}=\sin 20^\circ,\sin 40^\circ,\sin(-80^\circ)$$
since
$$\sin(3\times 20^\circ)=\sin(3\times 40^\circ)=\sin(3\times(-80^\circ))=\frac{\sqrt 3}{2}$$
with
$$\sin(-80^\circ)\lt \sin 20^\circ\lt \sin 40^\circ$$
Therefore, 
$$\color{red}{x=2\sin 20^\circ,\quad 2\sin 40^\circ,\quad -2\sin(80^\circ)}$$
