1. **State the problem:** We need to find the number of solutions (members) to the equation $$4 \sin 5\theta = 3$$ where $$\theta \in [0, 3\pi]$$. 2. **Rewrite the equation:** Divide both sides by 4 to isolate $$\sin 5\theta$$: $$\sin 5\theta = \frac{3}{4}$$ 3. **Recall the range and periodicity:** The sine function ranges between -1 and 1, so $$\frac{3}{4} = 0.75$$ is valid. 4. **General solution for $$\sin x = a$$:** $$x = \arcsin(a) + 2k\pi \quad \text{or} \quad x = \pi - \arcsin(a) + 2k\pi$$ for any integer $$k$$. 5. **Apply to our variable:** Let $$x = 5\theta$$, so $$5\theta = \arcsin\left(\frac{3}{4}\right) + 2k\pi \quad \text{or} \quad 5\theta = \pi - \arcsin\left(\frac{3}{4}\right) + 2k\pi$$ 6. **Find $$\arcsin(3/4)$$:** $$\arcsin\left(\frac{3}{4}\right) \approx 0.8481$$ radians. 7. **Express $$\theta$$:** $$\theta = \frac{0.8481 + 2k\pi}{5} \quad \text{or} \quad \theta = \frac{\pi - 0.8481 + 2k\pi}{5}$$ 8. **Determine valid $$k$$ values:** Since $$\theta \in [0, 3\pi]$$, For the first set: $$0 \leq \frac{0.8481 + 2k\pi}{5} \leq 3\pi$$ Multiply all sides by 5: $$0 \leq 0.8481 + 2k\pi \leq 15\pi$$ Subtract 0.8481: $$-0.8481 \leq 2k\pi \leq 15\pi - 0.8481$$ Divide by $$2\pi$$: $$\frac{-0.8481}{2\pi} \leq k \leq \frac{15\pi - 0.8481}{2\pi}$$ Calculate bounds: $$-0.135 \leq k \leq 7.44$$ So $$k = 0,1,2,3,4,5,6,7$$ (8 integer values). For the second set: $$0 \leq \frac{\pi - 0.8481 + 2k\pi}{5} \leq 3\pi$$ Multiply all sides by 5: $$0 \leq \pi - 0.8481 + 2k\pi \leq 15\pi$$ Subtract $$\pi - 0.8481$$: $$-(\pi - 0.8481) \leq 2k\pi \leq 15\pi - (\pi - 0.8481)$$ Calculate: $$-2.293 \leq 2k\pi \leq 14.848$$ Divide by $$2\pi$$: $$-0.365 \leq k \leq 2.36$$ So $$k = 0,1,2$$ (3 integer values). 9. **Count total solutions:** Number of solutions = 8 (first set) + 3 (second set) = 11. **Final answer:** There are **11** members in the set of roots for $$\theta \in [0, 3\pi]$$ satisfying $$4 \sin 5\theta = 3$$.