1. The problem is to determine if the function $\sin(x)$ is bijective. 2. A function is bijective if it is both injective (one-to-one) and surjective (onto). 3. The sine function $\sin(x)$ is periodic with period $2\pi$, meaning $\sin(x) = \sin(x + 2\pi k)$ for any integer $k$. 4. Because of this periodicity, $\sin(x)$ is not injective over all real numbers since multiple inputs produce the same output. 5. However, if we restrict the domain to $[-\frac{\pi}{2}, \frac{\pi}{2}]$, $\sin(x)$ is strictly increasing and covers the range $[-1,1]$. 6. On this restricted domain, $\sin(x)$ is injective and surjective onto $[-1,1]$, hence bijective. 7. Therefore, $\sin(x)$ is not bijective on $\mathbb{R}$ but is bijective when restricted to $[-\frac{\pi}{2}, \frac{\pi}{2}]$ with codomain $[-1,1]$. Final answer: $\sin(x)$ is bijective only on the restricted domain $[-\frac{\pi}{2}, \frac{\pi}{2}]$.