Are the twin primes the only positive double zeros of this real function?

For $$x \in \mathbb{R}, x \ge 1$$ define $$f(x) = \sin\frac{\pi(\Gamma(x)+1)}{\lfloor x \rfloor}$$

By Wilson's theorem the positive integer zeros of $$f(x)$$ are exactly the primes and the positive real zeros are when $$\Gamma(x)+1$$ is an integer multiple of $$\lfloor x \rfloor$$.

One can replace $$\lfloor x \rfloor$$ by $$x - \frac{i \, \log\left(-e^{\left(-2 i \, \pi x\right)}\right)}{2 \, \pi} - \frac{1}{2}$$ or $$x + \frac{\arctan\left(\cot\left(\pi x\right)\right)}{\pi} - \frac{1}{2}$$.

Consider $$t(x) = f(x) f(x+2)$$

Double positive integer zeros of $$t(x)$$ appear to be the smaller twin of twin primes, though the derivative of floor at integers complicate things.

Q1 Is it true that the smaller twin is a double zero of $$t(x)$$ (possibly with some relaxations of the definition)?

Q2 Does $$t(x)$$ have positive double real zeros which are not integers?

This might happen when $$\Gamma(x)+1$$ is an integer multiple of $$\lfloor x \rfloor$$,

$$\Gamma(x+2)+1$$ is an integer multiple of $$\lfloor x + 2\rfloor$$ and by the functional equation $$x (x + 1) \in \mathbb{Q}$$, i.e. $$\Gamma$$ must be integer at algebraic numbers which are not integers.

This approach might be extended to multiple zeros of prime tuples or both $$x,x^2+1$$ are primes.

Added A plot suggest counterexamples might exist, though couldn't find a counterexample so far neither by root finding nor by inverting gamma (modulo errors).

Let $$P(x)=\sin^2{(\pi x)} + \sin^2{( \pi(\Gamma(x)+1) / x)}$$
$$P(x)$$ vanishes exactly at the primes for real $$x$$ and is nonnegative.
So $$TP(x) = P(x) + P(x+2)$$ vanishes exactly at the smaller twin.
$$P(x)$$ has complex zeros.