1. The problem is to find how many solutions satisfy the equation $\sin x = x$ in the interval $[0, \pi]$. 2. We use the fact that $\sin x$ is a trigonometric function oscillating between -1 and 1, and $x$ is a linear function increasing from 0 to $\pi \approx 3.14159$. 3. At $x=0$, $\sin 0 = 0$ and $x=0$, so they are equal, giving one solution. 4. For $x > 0$, since $\sin x \leq 1$ and $x$ increases beyond 1, the line $y=x$ will be greater than $\sin x$ for all $x > 0$. 5. Therefore, the only solution in $[0, \pi]$ is at $x=0$. 6. Hence, the number of solutions to $\sin x = x$ in $[0, \pi]$ is 1.