<s>[INST] Solve the equation $x^{3}-3 x=\sqrt{x+2}$ 
Solve the equation $x^{3}-3 x=\sqrt{x+2}$

In the solution of this author wrote -
Let $-2 \leq x \leq 2 .$ Now Setting $x=2 \cos a, 0 \leq a \leq \pi$....
But here i did not understand the reason behind taking $ 0 \leq a \leq \pi$,i mean if we can take $a$ to be any real number since domain of $\cos x$ is real numbers then why we are restricting to
$ 0 \leq a \leq \pi$ ?
 [/INST] In the answer by Dan Fulea, there is a cubic factor  as "Just as a digression," namely $x^3 + x^2 - 2x - 1.$  This is well known, the roots are real numbers
$$ 2 \cos \left( \frac{2 \pi}{7} \right) \; \; , \; \;
2 \cos \left( \frac{4 \pi}{7} \right) \; \; , \; \;
2 \cos \left( \frac{8 \pi}{7} \right) \; \; . \; \; $$
This is verified easily enough, using nothing worse than the formula for $\cos 3 \theta. $
In the original problem $x^3 - 3 x = \sqrt{x+2},$ just one of the three works, namely $ x =2 \cos \left( \frac{4 \pi}{7} \right) \approx -0.445$
Note that the given roots are all strictly between $-2$ and $2.$
This page from Reuschle(1875) is missing from the scanned version online, I don't know what happened:


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