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 $\sin(20^\circ)$?

Hint:
When the cubic equation has $3$ real roots, you can use a trigonometric method: 
Setting $x=A\cos \theta$, we get the equation
$$A^3\cos^3\theta-3A\cos\theta+\sqrt3=A()+\sqrt3.$$
Now choose $A>0$ so that $A^2\cos^3\theta-3\cos\theta$ be proportional (even equal, here) to $4\cos^3\theta-3\cos\theta=\cos 3\theta$. We find $A=2$. The equation becomes
$$2\cos3\theta+\sqrt3=0\iff\cos3\theta=-\frac{\sqrt3}2.$$
The general solutions are
$$3\theta\equiv\pm\frac{5\pi}6\mod2\pi\iff\theta\equiv\pm\frac{5\pi}{18}\mod\frac{2\pi}3.$$
You can check this yields only $3$ different values for $\cos\theta$.
