Let the function $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \frac{\sin x}{e^{\pi x}} \left( \frac{x^{2023} + 2024x + 2025}{x^{2} - x + 3} \right) + \frac{2}{e^{\pi x}} \left( \frac{x^{2023} + 2024x + 2025}{x^{2} - x + 3} \right)$ Then the number of solutions of $f(x) = 0$ in $\mathbb{R}$ is ______.

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

$\text{Number of solutions of } f(x) = 0 \text{ in } \mathbb{R}$

$1.\ \text{write } f \text{ as a single fraction}$

$2.\ \text{sign of the denominator}$

$3.\ \text{range of } \sin x + 2, \text{ then count roots}$

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

$f(x) = 0 \iff (\sin x + 2)\,P(x) = 0$

$e^{\pi x} > 0, \qquad x^{2} - x + 3 = \left(x - \tfrac12\right)^{2} + \tfrac{11}{4} > 0$

$-1 \le \sin x \le 1 \ \Rightarrow\ \sin x + 2 \ge 1 > 0$

$f(x) = 0 \iff P(x) = 0$

$P(-1) = -1 - 2024 + 2025 = 0$

$P'(x) = 2023x^{2022} + 2024 > 0$

$\Rightarrow P \text{ strictly increasing} \Rightarrow \text{exactly one real root}$

$\Rightarrow \text{number of solutions} = 1$