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$$ x^3 - 3x + \sqrt{3} = 0 $$ 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 $\sin(20^o)$.

If not the cubic, please tell me how to evaluate $\sin(20^o)$?

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    The cubic has two positive roots and one negative root. – 2017-02-28
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    Yes, but how to solve – 2017-02-28
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    The roots will have $\sqrt{-1}$. But once you approximate it, all the ugly $\sqrt{-1}$ cancel out. – 2017-02-28
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    Actually, it's a feature of Cardano's formula: when a cubic equation has real roots, the computation with this formula involves complex numbers that will cancel out. – 2017-02-28
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    @zipirovich - Please tell me how to cancel it out. – 2017-02-28
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    See http://math.stackexchange.com/questions/2157643/how-can-i-solve-the-equation-x3-x-1-0/2157645#2157645 – 2017-02-28

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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, $$\color{red}{x=2\sin 20^\circ,\quad 2\sin 40^\circ,\quad -2\sin(80^\circ)}$$

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    NOOOOOO!!!!! You took the answer I was going to put... (+1) – 2017-02-28
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    Is the Cardano's Method a failure in this case and, as mentioned in the question, I encountered the cubic while trying to evaluate sin(20 degree) and, in your solution we come back to it so, is there no way to calculate sin(20 degree) – 2017-02-28
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    I thought the OPs question was essentially "I know $\sin(20^\circ)$ is a root of the polynomial, but I'm trying to solve the polynomial in radicals to find a radical form for $\sin(20^\circ)$". Maybe I am misreading the OP... – 2017-02-28
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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 $\cos\theta$.

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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.