Solve $x^{3}-3x=\sqrt{x+2}$ Solve for real $x$   
$$x^{3}-3x=\sqrt{x+2}$$   

By inspection, $x=2$ is a root of this equation. So, I squared both sides and divided the six degree polynomial obtained by $x-2$. Then I got a quintic which I couldn't solve despite applying rational root theorem and substitutions. I believe that there must be some nice method to solve this which I can't think about. Please help. Thanks!


Two preliminary remarks.
If $x$ is solution, then $x(x+\sqrt{3})(x-\sqrt{3}) = x^3-3x \ge 0$, so $-\sqrt{3} \le x \le 0$ or $x \ge \sqrt{3}$.
Observe also that $2$ is solution and no real number larger than $2$ can be solution since for $\ge 2$,
$$\frac{d}{dx}(x^3-3x) = 3x^2-3 \ge 9 > \frac{1}{4} \ge \frac{1}{2\sqrt{x+2}} = \frac{d}{dx}(\sqrt{x+2}).$$
So let $x \in [-2,2]$. Introduce the unique $t \in [0,\pi/2]$ such that $x = 2\cos{2t}$ with $t \in [0,\pi/2]$.
\begin{eqnarray*}
x^3-3x = \sqrt{x+2} 
&\iff& 8\cos^3{2t}-6\cos2t = \sqrt{(2\cos2t+2)} \\
&\iff& 2\cos(6t) = 2\cos(t) \textrm{ since } \cos(t) \ge 0 \\
&\iff& 6t \pm t \in 2\pi\mathbb{Z} \\
&\iff& t \in \{0,2\pi/7,22\pi/5\} \textrm{ since } t \in [0,\pi/2] \\
&\iff& X \in \{2,2\cos(2\pi/7),2\cos(2\pi/5)\}.
\end{eqnarray*}
