Question:

Step 1: Determine the range of the innermost function $\sin x$. For all $x \in \mathbb{R}$, we know that: $$\sin x \in [-1, 1]$$ Step 2: Find the range of the argument $\frac{\pi}{2}\sin x$. Since $\sin x \in [-1, 1]$, multiplying by $\frac{\pi}{2}$ gives: $$\frac{\pi}{2}\sin x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ Step 3: Determine the range of $\sin\left(\frac{\pi}{2}\sin x\right)$. Since $\frac{\pi}{2}\sin x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and the sine function is continuous and monotonically increasing on this interval, we have: $$\sin\left(\frac{\pi}{2}\sin x\right) \in [-1, 1]$$ Step 4: Find the range of the argument $\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)$. Since $\sin\left(\frac{\pi}{2}\sin x\right) \in [-1, 1]$, multiplying by $\frac{\pi}{6}$ gives: $$\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right) \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$$ Step 5: Determine the range of $f(x) = \sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\right)$. Since $\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right) \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$ and the sine function is continuous and monotonically increasing on this interval, we have: $$f(x) = \sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\right) \in \left[-\sin\left(\frac{\pi}{6}\right), \sin\left(\frac{\pi}{6}\right)\right]$$ Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, we get: $$f(x) \in \left[-\frac{1}{2}, \frac{1}{2}\right] = [-0.5, 0.5]$$ **Final Answer:** The range of $f(x)$ is $\boxed{[-0.5, 0.5]}$, which corresponds to **Option 3**.