Let the function $$f : \mathbb{R} \to \mathbb{R}$$ be defined by

$$f(x) = \frac{\sin x}{e^{\pi x}} \cdot \frac{(x^{2023} + 2024x + 2025)}{(x^2 - x + 3)} + \frac{2}{e^{\pi x}} \cdot \frac{(x^{2023} + 2024x + 2025)}{(x^2 - x + 3)}$$.

Then the number of solutions of $$f(x) = 0$$ in $$\mathbb{R}$$ is ______.

First, rewrite the given function in a factored form.

$$f(x)=\frac{\sin x}{e^{\pi x}}\cdot\frac{x^{2023}+2024x+2025}{x^{2}-x+3}+\frac{2}{e^{\pi x}}\cdot\frac{x^{2023}+2024x+2025}{x^{2}-x+3}$$

Factor out the common terms $$\dfrac{x^{2023}+2024x+2025}{x^{2}-x+3}\cdot e^{-\pi x}$$:

$$f(x)=\frac{x^{2023}+2024x+2025}{x^{2}-x+3}\,e^{-\pi x}\,(\sin x+2)$$

To solve $$f(x)=0$$ we must examine when each factor can be zero.

1. The exponential factor $$e^{-\pi x}$$ is never zero for any real $$x$$.
2. The quadratic denominator $$x^{2}-x+3$$ has discriminant $$(-1)^{2}-4(1)(3)=-11\lt 0$$, so it is always positive and never zero.
3. The factor $$\sin x+2$$ cannot be zero because $$\sin x\in[-1,1]$$, so $$\sin x+2\in[1,3]$$ for all $$x\in\mathbb{R}$$.

Therefore the only possible zeros come from

$$x^{2023}+2024x+2025=0$$

Define $$g(x)=x^{2023}+2024x+2025$$. To count its real roots, study monotonicity:

$$g'(x)=2023x^{2022}+2024$$

Since $$x^{2022}\ge 0$$ for every real $$x$$, we have $$g'(x)\ge 2024\gt 0$$. Thus $$g(x)$$ is strictly increasing on the entire real line.

A strictly increasing odd-degree polynomial with positive leading coefficient satisfies

$$\lim_{x\to-\infty}g(x)=-\infty,\qquad \lim_{x\to+\infty}g(x)=+\infty$$

By the Intermediate Value Theorem, it crosses the $$x$$-axis exactly once. Hence $$g(x)=0$$ has exactly one real solution.

Because no other factor of $$f(x)$$ can become zero, this single root of $$g(x)$$ is the only root of $$f(x)$$.

Therefore, the number of solutions of $$f(x)=0$$ in $$\mathbb{R}$$ is

1