Determine (approximately) the interval for x on which
Determine (approximately) the interval for $x$ on which the approximation $\sin x\approx x$ holds with an accuracy of $1\%$ of the value of $\sin x$.
Explanation of Method
A first-order estimate of the relative error gives a clean threshold that only slightly overshoots the exact one.
Step 1: Set up the relative-error condition
For small $x>0$, $\sin x\approx x$, so the condition $\dfrac{|\sin x-x|}{|\sin x|}\lt 0.01$ becomes
\[ \frac{x-\sin x}{\sin x} \lt 0.01. \tag{1} \]
Step 2: Leading-order estimate
From $\sin x=x-\tfrac{x^3}6+\cdots$, the numerator is $x-\sin x\approx\tfrac{x^3}6$; approximating the denominator by $\sin x\approx x$ as well, (1) becomes
\[ \frac{x^3/6}{x} \lt 0.01 \;\Longrightarrow\; \frac{x^2}6 \lt 0.01 \;\Longrightarrow\; x^2\lt 0.06 \;\Longrightarrow\; x \lt \sqrt{0.06} = \frac{\sqrt6}{10} \approx 0.245. \tag{2} \]
Step 3: Verification
\[x=0.24\text{: relative error}=(x-\sin x)/\sin x\approx0.00966\lt 0.01\ \checkmark\]
\[x=\sqrt{0.06}\approx0.24495\text{: relative error}\approx0.01007,\ \text{slightly over }1\%;\ \text{bisecting numerically, the true threshold is }x^*\approx0.24410\]
Result
Final answer:
\[\boxed{\,|x| < \sqrt{0.06} \approx 0.245\ \text{rad} \approx 14^\circ\,}\]
Remark
Remark. The boxed value $\sqrt{0.06}$ comes from replacing $\sin x$ by $x$ in both the numerator's leading term and the denominator — a first-order approximation twice over. A direct numerical bisection of the exact condition $(x-\sin x)/\sin x=0.01$ gives the true threshold $x^*\approx0.2441$, about $0.3\%$ smaller than $\sqrt{0.06}\approx0.2449$; at $\sqrt{0.06}$ itself the relative error is actually $\approx1.007\%$, a hair over the target. For a result meant to be "approximate" (as the problem asks), $\sqrt{0.06}\approx0.245$ is the standard and expected leading-order answer; Problem 623 below repeats the same calculation with one more term of $\sin x$ kept, and there the resulting bound matches its own verification essentially exactly.