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 $\sin(20^\circ)$?
 A: 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, 
$$\color{red}{x=2\sin 20^\circ,\quad 2\sin 40^\circ,\quad -2\sin(80^\circ)}$$
A: As @mathlove points out, $2\sin20^\circ$ is the smaller positive root of the equation. If you use Cardano's method, you will end up with real roots, although via cube roots of complex numbers, which are inconvenient to calculate if you can't use inverse trigonometric functions. To get a good numerical approximation, the Newton–Raphson method is very efficient.
A: 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 $\cos\theta$.
A: If what you want is an expression for the sine of $20°$ with real radicals, there is a minor problem: it does not exist.
Immovable object
You may want to read up on the casus irreducibilis, a situation that emerges when you try to solve certain cubic equations by the general method to obtain solutions with radicals. Whenever the equation has three real roots, the cube root radicals come out with complex arguments, and the only way to combine them into a real value is to revert to the equation you attempted to solve. The WP article referenced above contains a "simple" proof, but "simple" means it can be rendered within a few lines of text. It requires knowledge of Galois theory to comprehend.
We can get around the casus irreducibilis by using trigonometric functions, but this is of limited utility if your objective is to simplify trigonometric functions in the first place. With your problem of rendering $\sin 20°$ you get the following enlightened result:
$\sin 20°=\sin 20°$
or something equally informative like $\cos 70°$. In some other cases you end up invoking other trigonometric functions in place of the one you tried to simplify. Solving $x^3+x^2-2x-1=0$ to find $2\cos(360°/7)$ gets you a solution involving $\cos[(1/3)\arctan(\sqrt{27/28})]$.
And it isn't just cubic equations. The term "casus irreducibilis" is most often used for degree 3, but similar things happen in irreducible higher degree polynomial equations that appear to be solved by radicals but have more than one real root. For instance, trigonometric functions involving $360°/11$ can be expressed as roots of quintic equations, the quintic equations can be solved by radicals ... but the radicals are complex and can't be resolved without reverting to the quintic equation.
Euclid's irresistible (but limited) force
Only square roots of complex numbers can generally be resolved to get their separate real and imaginary parts, so the trigonometric functions you can actually get in terms of real radicals are those where the radicals are all square roots. Meaning, in geometric terms, the angles can be constructed using unmarked straightedge and compasses. Thus you can construct $18°$ and the sine of that angle, $(\sqrt5-1)/4$, presents no casus irreducibilis problems. The famous finding that the regular 17-gon is constructible similarly comes with real square-root expressions for trigonometric functions of multiples of $360°/17$.
