Why can we let $x = 2\cos t\ $ in the solution for the following system of equations

Question

Solve in real number the system of equations $\begin{cases}x^2 = y+2 \\ y^2 = z+2 \\ z^2 = x+2 \end{cases}$

The solution given to me says the following:

If we eliminate $y$ and $z$, we obtain a polynomial, $P$, of degree $8$ in $x$. Clearly this not an efficient way to proceed. Let $x = 2\cos t,\ 0 \leq t \leq \pi$...

Why are we allowed to make this trignometric substitution? To me, this is saying I can represent any real number $x$ by the trig function $2\cos t$ on the domain $t \in \left [ 0, \pi\right]$ because $x$ is supposed to be any real number. But, $|2\cos t|$ is at most $2$ on that interval. So why are we allowed to do this substitution if I cannot represent every real number with that trig function?

Answer 1

Because $x\geq-2$, $y\geq-2$ and $z\geq-2$ and if $x>2$ so $z^2>2+2=4,$ which gives $z>2$ and from here $y>2,$ which is a contradiction because $$6=\sum_{cyc}(x^2-x)>6.$$ We used that $x^2-x$ increases for $x>2$.

Id est, we got $$\{x,y,z\}\subset[-2,2].$$ Also, $\cos$ decreases on $[0,\pi]$, which says that the substitution $x=2\cos{t}$ is correct.

Answer 2

I can't resist writing this beautiful solution.

Setting $x=2\cos t$,

$$x^2=y+2\implies 4\cos^2t-2=y\implies y=2\cos2t.$$

Then by the same principle,

$$z=2\cos4t$$

and

$$x=2\cos 8t.$$

Finally, the solutions are the roots of

$$\cos8t=\cos t.$$

To avoid duplicates, we only search in the range $t\in[0,\pi)$, and write

$$8t=\pm t+2k\pi.$$

This gives $8$ distinct real solutions of the form $x=2\cos\dfrac{2k\pi}7$ and $2\cos\dfrac{2k\pi}9$, which is the maximum we can expect from an octic problem.

This validates the change of variable a posteriori.