Number of Real Solutions of Trigonometric Rational Expression Equation

To find the number of real solutions of the equation $f(x) = 0$, we start by simplifying the given expression for $f(x)$:

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

Factoring out the common terms, we get: $f(x) = \frac{\left( x^{2023} + 2024x + 2025 \right)(\sin x + 2)}{e^{\pi x} \left( x^2 - x + 3 \right)}$

Now, let us analyze each component of $f(x)$ for $x \in \mathbb{R}$:

1. Exponential Term ($e^{\pi x}$): $e^{\pi x} > 0 \quad \forall x \in \mathbb{R}$

2. Quadratic Term ($x^2 - x + 3$): The discriminant of $x^2 - x + 3$ is: $\Delta = (-1)^2 - 4(1)(3) = 1 - 12 = -11 < 0$ Since the leading coefficient is $1 > 0$, we have: $x^2 - x + 3 > 0 \quad \forall x \in \mathbb{R}$

3. Trigonometric Term ($\sin x + 2$): Since $-1 \le \sin x \le 1$ for all $x \in \mathbb{R}$, adding $2$ gives: $1 \le \sin x + 2 \le 3 \implies \sin x + 2 > 0 \quad \forall x \in \mathbb{R}$

Since the denominator $e^{\pi x}(x^2 - x + 3)$ is strictly positive and non-zero for all real $x$, and the factor $(\sin x + 2)$ is also strictly positive for all real $x$, the equation $f(x) = 0$ simplifies to setting the remaining polynomial factor to zero:

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

Let $P(x) = x^{2023} + 2024x + 2025$. Differentiating $P(x)$ with respect to $x$: $P'(x) = 2023 x^{2022} + 2024$

Since $x^{2022} \ge 0$ for all $x \in \mathbb{R}$, we have: $P'(x) \ge 2024 > 0 \quad \forall x \in \mathbb{R}$

Since $P'(x) > 0$ for all $x \in \mathbb{R}$, $P(x)$ is a strictly increasing continuous function on $\mathbb{R}$.

Furthermore, $P(x)$ is an odd-degree polynomial, so: $\lim_{x \to -\infty} P(x) = -\infty \quad \text{and} \quad \lim_{x \to \infty} P(x) = \infty$

By the Intermediate Value Theorem and strict monotonicity, $P(x) = 0$ has exactly one real solution.

Thus, the number of solutions of $f(x) = 0$ in $\mathbb{R}$ is 1.