We are given the function $f(x) = \sin(x) + \cos(x)$ and tasked with finding the smallest positive integer $X$ such that the function $g(x) = f(x) \cdot f(x+X)$ has a period of $\pi$. Additionally, we are provided that the multiplicity of the solution to the equation $f(x) = 0$ in the interval $[0, 2\pi)$ is $m$, and that $\forall x \in [0, 2\pi), \exists y \in [0, 2\pi)$ such that $f(x) \cdot f(y) = 0$.

Step 1: Simplify $f(x)$

We start by simplifying the function $f(x) = \sin(x) + \cos(x)$.

Using the trigonometric identity $\sin(x) + \cos(x) = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$, we can express $f(x)$ in terms of a single sine function: $f(x) = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$ Thus, the function $f(x)$ is simply a sine wave, shifted by $\frac{\pi}{4}$, and its zeros correspond to the zeros of $\sin(x)$.

Step 2: Find the zeros of $f(x)$

The function $f(x) = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$ has zeros where the sine function is zero. That is, we need to solve $\sin\left(x + \frac{\pi}{4}\right) = 0$ The general solution to this equation is: $x + \frac{\pi}{4} = n\pi \quad \text{for some integer } n$ Thus, $x = n\pi - \frac{\pi}{4}$ For $x \in [0, 2\pi)$, the solutions are: $x = \frac{3\pi}{4}, \frac{7\pi}{4}$ Therefore, the zeros of $f(x)$ in $[0, 2\pi)$ are $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$.

Step 3: Analyze $g(x) = f(x) \cdot f(x+X)$

We are tasked with finding the smallest positive integer $X$ such that $g(x) = f(x) \cdot f(x+X)$ has a period of $\pi$.

First, observe that $f(x)$ itself has a period of $2\pi$, since $\sin(x + \frac{\pi}{4})$ is periodic with period $2\pi$.

Now, let's analyze the function $f(x+X)$. The zeros of $f(x+X)$ will occur when $x+X$ satisfies the same zero conditions as $f(x)$, namely: $\sin\left((x+X) + \frac{\pi}{4}\right) = 0$ The key idea is that for $g(x) = f(x) \cdot f(x+X)$ to have a period of $\pi$, the zeros of $f(x)$ and $f(x+X)$ must be aligned in such a way that $g(x)$ repeats every $\pi$.

Step 4: Determine the smallest $X$

Since $f(x)$ has zeros at $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$, the function $f(x+X)$ must have zeros that interlace with the zeros of $f(x)$, leading to $g(x)$ repeating every $\pi$.

The smallest such $X$ occurs when the shift aligns $f(x+X)$ with $f(x)$ such that the period of $g(x)$ becomes $\pi$. This happens when $X = \pi$, because shifting by $\pi$ causes the sine function to "flip," resulting in the desired periodic behavior.

Step 5: Verify the multiplicity and conditions

The multiplicity $m$ of the solution $f(x) = 0$ in $[0, 2\pi)$ is $2$ because there are two distinct zeros, $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$.

The condition that $\forall x \in [0, 2\pi), \exists y \in [0, 2\pi)$ such that $f(x) \cdot f(y) = 0$ is satisfied because for any $x$, there exists a zero of $f(y)$ within $[0, 2\pi)$.

Final Answer:

The smallest positive integer $X$ is $\boxed{1}$.