1. **State the problem:** We are given the function $s(x) = \mu[\sin(2x) - 1]$ defined on the interval $[-\pi, \pi]$ and the inequality $s(x) < \frac{1}{2}$. We need to determine the set $A_s(x)$ where this inequality holds. 2. **Rewrite the inequality:** $$\mu[\sin(2x) - 1] < \frac{1}{2}$$ 3. **Isolate the sine term:** $$\sin(2x) - 1 < \frac{1}{2\mu}$$ 4. **Add 1 to both sides:** $$\sin(2x) < 1 + \frac{1}{2\mu}$$ 5. **Analyze the range:** Since $\sin(2x)$ ranges between $-1$ and $1$, the inequality depends on the value of $1 + \frac{1}{2\mu}$. - If $1 + \frac{1}{2\mu} \geq 1$, then the inequality $\sin(2x) <$ something $\geq 1$ is always true because $\sin(2x) \leq 1$. - If $1 + \frac{1}{2\mu} < -1$, then no $x$ satisfies the inequality. - Otherwise, we solve for $x$ where $\sin(2x) < 1 + \frac{1}{2\mu}$. 6. **Assuming $1 + \frac{1}{2\mu} \in (-1,1)$, solve:** Let $c = 1 + \frac{1}{2\mu}$. We want: $$\sin(2x) < c$$ 7. **Find the general solution:** The solutions to $\sin(2x) = c$ are: $$2x = \arcsin(c) + 2k\pi \quad \text{or} \quad 2x = \pi - \arcsin(c) + 2k\pi, \quad k \in \mathbb{Z}$$ Since $\sin(2x)$ is increasing on $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the inequality $\sin(2x) < c$ holds for: $$2x \in (-\frac{\pi}{2} + 2k\pi, \arcsin(c) + 2k\pi) \cup (\pi - \arcsin(c) + 2k\pi, \frac{3\pi}{2} + 2k\pi)$$ 8. **Restrict to $x \in [-\pi, \pi]$:** Divide the intervals by 2: $$x \in \left(-\frac{\pi}{4} + k\pi, \frac{\arcsin(c)}{2} + k\pi\right) \cup \left(\frac{\pi}{2} - \frac{\arcsin(c)}{2} + k\pi, \frac{3\pi}{4} + k\pi\right)$$ For $k = -1, 0$ to cover $[-\pi, \pi]$. 9. **Final answer:** $$A_s(x) = \bigcup_{k=-1}^0 \left(\left(-\frac{\pi}{4} + k\pi, \frac{\arcsin\left(1 + \frac{1}{2\mu}\right)}{2} + k\pi\right) \cup \left(\frac{\pi}{2} - \frac{\arcsin\left(1 + \frac{1}{2\mu}\right)}{2} + k\pi, \frac{3\pi}{4} + k\pi\right)\right) \cap [-\pi, \pi]$$ This set describes all $x$ in $[-\pi, \pi]$ where $s(x) < \frac{1}{2}$. If $1 + \frac{1}{2\mu} \geq 1$, then $A_s(x) = [-\pi, \pi]$. If $1 + \frac{1}{2\mu} \leq -1$, then $A_s(x) = \emptyset$.