1. We want to find the range of $x$ values for which the error in approximating $\sin x$ by $x - \frac{x^3}{6}$ is at most $5 \times 10^{-4}$. 2. The Taylor series of $\sin x$ about 0 is $x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$. 3. The error after the cubic term can be estimated by the next term in the series: $\left|\frac{x^5}{120}\right|$. 4. We want $\left|\frac{x^5}{120}\right| \leq 5 \times 10^{-4}$. 5. Multiply both sides by 120: $|x|^5 \leq 120 \times 5 \times 10^{-4} = 0.06$. 6. Take the fifth root: $|x| \leq (0.06)^{\frac{1}{5}}$. 7. Calculating $(0.06)^{\frac{1}{5}}$ gives approximately $0.549$ 8. Therefore, the approximation $\sin x \approx x - \frac{x^3}{6}$ holds with an error not greater than $5 \times 10^{-4}$ when $|x| \leq 0.549$. **Final answer:** $$|x| \leq 0.549$$