I will give a high level discussion of my perspective on this type of
equation-solving and what is and isn't logically valid. My discussion will not
necessarily be suited for the average A-level student (and I agree with what you
say - from my recollection of taking A-levels, I think the exam board likely doesn't actually expect much justification
or logical rigour here).
In general, in the process of "solving an equation $f(x) = 0$" (ie finding an
explicit description for the set $X$ of $x$ such that $f(x) = 0$), one technique
is to repeatedly "do something to both sides" of the equality and apply
algebraic identities, ending up with some equation $g(x) = 0$ which is easier to
analyse. Say we know that $X'$ is exactly the set of $x$ such that $g(x) = 0$.
Since we manipulated the original equation by a series of "$\implies$" steps, we
know that $X \subseteq X'$. The step that students often get wrong is jumping
the gun and assuming that $X = X'$. But to be fully rigorous at this stage, one
must prove this - ie check that $X' \subseteq X$. It's quite possible that if
you've been overly zealous in your manipulations, you've overshot and $X$ might
be a proper subset of $X'$. If $X'$ is still a nice enough set, you can still
find $X$ by just taking the subset of $X'$ of $x$ that do satisfy $f(x) = 0$.
The classic example is an equation like $\sqrt{|x + 1|} = x$. If you square both
sides, you get $|x + 1| = x^2$, which you can solve to see
$x = \tfrac 12 (1 \pm \sqrt 5)$. But of course one of these is not actually a
solution.
In this case, the template of the solution is quite different. We wish to solve
$f(x) = 0$. The way we do this is to consider some function
$c: [0, \pi] \to [-1, 1]$, and instead we look for solutions to
$f(c(\theta)) = 0$. If $\Theta$ is the set of such $\theta$, then
$\{c(\theta) : \theta \in \Theta\} \subseteq X$. However, it is possible
that it's a proper subset. This means that there may be cases where we carry out
this argument and find a small number of solutions, but one cannot conclude
immediately that those are all the solutions. If you're aware of this, then
there's no more logical peril than in the other template I describe - you just
have to be careful to verify that you've not undershot and have indeed found all
the solutions.
In the case of your particular question, one can verify that (with
multiplicities) the three roots obtained are all the roots of your equation, by
using the theorem that a quintic has at most $5$ roots with multiplicity. Since
what you've written doesn't include any justification that you have the full set
of solutions, I would agree that you don't know yet if you've found all the
roots to the equation. But if you do this verification, I would say that
solution is on logically solid ground. However, it does get a bit "lucky", and
there are similar problems for which that technique would fail to solve the
problem.
(Incidentally, there are all sorts of ways to verify that you have actually got
all the roots. Angelo's answer gives one way to see it. You can also look at
derivatives. I think a nice way is to
argue that for $k$ a tiny bit less than $1$, the equation
$16x^{5}-20x^{3}+5x+k = 0$ does have $5$ distinct real roots, and as we take the
limit $k \to 1$, two of those pairs of roots merge together. Hence those must be
repeated roots.)
Here is an example of such a problem: say we're trying to solve
$16x^{5}-20x^{3}+5x+2 = 0$. Suppose $x = \cos \theta$ for
$\theta \in [0, \pi]$. Then by the nice identity we proved, we have
$\cos 5\theta + 2 = 0$, so $\cos 5\theta = -2$. This is absurd, because for
$\theta \in \Bbb R$, we have $\cos \theta \ge -1$. Therefore there are no
solutions obtained by this method. However, there is in fact a real solution
to this equation, which follows from the fact that this polynomial has odd
degree. This is frustrating!
We can actually make a small change to the second template to make it more
powerful. If the function $c$ is surjective onto the domain of $f$, then
actually we're guaranteed to have $\{c(\theta) : \theta \in \Theta\} = X$
(exercise for the reader). This is a very nice guarantee, particularly if the
equation $f(c(\theta)) = 0$ is easy to work with.
Luckily for us, the function $\cos$ extends to $\Bbb C$, to give a surjective map
$\Bbb C \to \Bbb C$. Moreover, the identity $\cos 5\theta \equiv 16 \cos^5
\theta - 20\cos^3 \theta + 5\cos \theta$ remains valid for $\theta \in \Bbb C$,
and so do many familiar facts about the behaviour of $\cos$ (these facts do all
need proving. If you're familiar
with complex analysis, this all follows from general nonsense like Picard's
theorem and the identity principle). So we will be able to use the same
technique to find all complex roots of $16x^{5}-20x^{3}+5x+2 = 0$, from which we can work out the real roots if we like. I will write
out this whole solution for the more advanced reader's benefit, but this is
probably not something the average A-level student would be expected to follow.
Off we go, then. Let $\theta \in \Bbb C$. We wish to solve the equation
$\cos 5\theta = -2$. Let's write $5\theta = u + iv$ with $u, v \in \Bbb R$.
Then by standard identities, we have
$\cos(u + iv) = \cos u \cosh v + i \sin u \sinh v$,
so we must have $\cos u \cosh v = -2$ and $\sin u \sinh v = 0$.
From the latter equation it follows that $v = 0$ or $u = \pi n$ for some $n \in \Bbb Z$.
But we already know that $v$ cannot be $0$.
Now the former equation says $(-1)^n \cosh v = -2$. Since $v$ is real, we have
$\cosh v \ge 0$. So $n$ must be odd and we must have
$\cosh v = 2$, and hence $v = \pm \operatorname{arccosh} 2$.
So, the general solution to the original equation is
$x = \cos(\tfrac 15((2n + 1)\pi \pm i \operatorname{arccosh} 2))$
(in other words,
$\Theta = \{\tfrac 15((2n + 1)\pi \pm i \operatorname{arccosh} 2) : n \in \Bbb Z\}$
and we are now looking at $\{\cos \theta : \theta \in \Theta\}$).
Here the choice of sign for the $\pm$ gives the conjugate pairs of solutions to
this quintic. You can work out the ugly formula in terms of $\cosh$ and $\sinh$
if you like.
Taking $n = 0$ gives the solutions
$0.837242435715400 \pm 0.156613988262999i$.
Taking $n = 1$ gives the solutions
$-0.319798153619534 \pm 0.253406756123210i$.
Taking $n = 3$ gives the real solution $-1.03488856419173$. Hurray!
Now we can argue that by periodicity and symmetry properties of $\cos$, this must be all of
$\{\cos \theta : \theta \in \Theta\}$, and hence all of $X$. Alternatively, we
already know that $|X| \le 5$ because $X$ is a quintic. So we've found all
solutions.
Lastly - if your solution had used the modified template, with $\theta \in \Bbb
C$, then you could actually be confident you had all the solutions without doing
an extra verification, so long as you had justified well enough that you've
found all possible $\theta$
(and so long as you accept that $\cos: \Bbb C \to \Bbb C$ is surjective). If you know enough about the complex cosine
function, you would in fact have to change almost nothing about the argument,
because the complex cosine function in general doesn't introduce "extra"
preimages of numbers in $[-1, 1]$. But again, such knowledge is not expected of
A-level students and it's not something they can assume in their answers to exam
questions, as far as I know.
(In fact more generally,
"$\cos x = \cos y$ iff $x - y \in 2\pi \Bbb Z$ or $x + y \in 2\pi \Bbb Z$"
still holds for $x, y \in \Bbb C$.
This is because
$\cos x - \cos y = -2\sin(\tfrac 12(x + y))\sin(\tfrac 12(x - y))$,
and $\sin z = 0$ iff $z \in \pi \Bbb Z$.
You can see this last fact by directly working it out,
or by some fun general nonsense
again :) - if there was some $z_0 \notin \Bbb R$ with $\sin z_0 = 0$, then
$2z_0$ would be a period of $\sin$, because
$\sin(z + 2z_0) = \sin z \cos 2z_0 + \cos z \sin 2z_0
= \sin z (1 - 2 \sin^2 z_0) + 0 = \sin z$,
but any doubly-periodic entire function must be constant.)
