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$ ?
 A: Note that for
$$0 \leq a \leq \pi \implies -1\le \cos a \le 1 \implies -2\le 2\cos a \le 2$$
which suffices to cover by a bijection the whole range for $x=2\cos a$.
Note that the assumption is due to these facts

*

*$\sqrt{x+2}$ requires $x\ge -2$

*for $x>2 \implies x^{3}-3 x>\sqrt{x+2}$
A: They've chosen the principal value branch of $\arccos$.
$-2\le 2\cos a \le 2 \Rightarrow -1 \le \cos a\le 1 \Rightarrow \pi\ge a \ge 0$ or $0\le a\le\pi$
A: It would be good as a first step to have an estimation for the solutions, else, without knowing that any potential real solution ($\ge -2$) lives in the interval $[-2,2]$, the argument is only partial. The question is of course valid.
Now, one possibility to work is to start with the given equation, and go in one direction to see what follows from it. We obtain:
$$(x^3-3x)^2-(x+2)=0\ .$$
Just as a digression, one can factorize the above as:
$$
(x-2)(x^2+x-1)(x^3 +x^2-2x-1)\ .
$$
For the first two factors we have "easy solutions". But there are some problems with the third factor. Digression finished, this will not be used, but it may be maybe compared finally with the obtained solutions.
At any rate, there are at most six solutions of the given polynomial equation of degree six.

Now we abandon this path, and try to get as many solutions as possible in the interval $[-2,2]$, this was the hint for question maybe.
(If we already find six solutions, we are done!)
So we are modest and search first only for such solutions. The substitution $x=2\cos 2t$ for some "angle" $t$ delivers immediately:
$$
(\dagger)
\qquad
2\cos 6t = 2|\cos t|\  .
$$
And from the weaker equation $\cos 6t \pm \cos t=0$ we collect six solutions for $(  \dagger)$ for $t$, which are
$$
0\ ,\ 
\frac {2\pi}7\ ,\ 
\frac {2\pi}5\ ,\ 
\frac {4\pi}7\ ,\ 
\frac {4\pi}5\ ,\ 
\frac {6\pi}7\ .
$$

The above discussion leads to my answer for the OP: We just do that substitution, thus searching only for solutions in $[-2,2]$. We expect - without any restriction - maximally six solution. A posteriori, after the trigonometric substitution, we have indeed six solutions. "So we had luck" and "the substitution works".

Final note: In fact, if we have the above list, we can check individually that each value in the list is a solution. And with the same argument, maximally six solutions, the problem is solved with minimal typing effort.
A: 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:


