As Timothy Chow pointed out, the OP's question is most naturally
formulated as a promise problem: given a finite system of
equations and inequalities in finitely many variables and the
promise that it has a solution and furthermore that all solutions
have the same value for the particular variable $\xi$, determine
whether or not that value of $\xi$ is rational.

Let me show here in various ways that if we drop the
promise-problem aspect of the problem, it is not decidable.

**Theorem.** There is no computable procedure to determine whether
a given finite system of equations and inequalities, allowing the
operations $+,\cdot,0,1,\sin$, has a solution in the reals.

**Proof.** We shall use the variables $\xi,x_1,\ldots,x_n$ and
another variable $p$. By insisting that $\sin(p)=0$ and $p>0$, we
can ensure that $p$ is an positive integer multiple of $\pi$. By
insisting that $\sin(x_i\cdot p)=0$, we can ensure that each $x_i$
is an integer.

In this way, we may transform any diophantine equation $q(\vec
x)=0$, where $q$ is polynomial in several variables over the
integers, into a system over the reals, such that $q(\vec x)=0$
has a solution over the integers if and only if the new system has
a solution over the reals.

But it is a well-known consequence of the MRDP theorem
that we cannot computably decide whether a given diophantine
equation has a solution over the integers (and this was the
solution to Hilbert's 10th problem). **QED**

**Theorem.** There is no computable procedure to determine whether
a given finite system of equations and inequalities in the
variables $x_1,\ldots,x_n,\xi$, allowing the operations
$+,\cdot,0,1,\sin$, has at least one rational value $\xi$ that is
part of a solution. Furthermore, this remains undecidable even
given the promise that there is at most one value $\xi$ that is
part of a solution, and that no irrational $\xi$ are solutions.

**Proof.** For any diophantine equation $q(\vec x)=0$, consider
the modified system over the reals, using the additional variable
$p$, where we insist that $q(\vec x)=0$ and also $\sin(p)=0$,
$p>0$, and $\sin(x_i\cdot p)=0$, which ensure that $p$ is a
positive integer multiple of $\pi$ and that all the $x_i$'s are
integers. Finally, add the equation $\xi=0$.

The original diophantine equation has a solution over the integers
just in case the new system has a solution over the reals, and in
any such solution, $\xi$ will be rational, because it will be $0$.
So the original system has a solution over the integers if and
only if the new system has at least one solution for which $\xi$
is rational. So this is not decidable. **QED**

**Theorem.** There is no computable procedure to determine whether
a given finite system of equations and inequalities in the
variables $x_1,\ldots,x_n,\xi$, allowing the operations
$+,\cdot,0,1,\sin$, has at most one rational $\xi$ that is part of
a solution. Furthermore, this remains undecidable under the
promise that there are at most two values of $\xi$ that solve the
system, both rational.

**Proof.** We use the same idea, but replace the equation $\xi=0$
with the equation $\xi(\xi-1)=0$. So this system has either no
solutions or $\xi=0$ and $\xi=1$ are both solutions. **QED**

These theorems do not seem to answer the full question: given a finite system of equations and inequalities, and given the promise that there is exactly one value of $\xi$ that is part of a solution, determine whether this value of $\xi$ is rational or not.