1. **State the problem:** Solve the trigonometric equation $$3 + 5 \sin \theta = 1$$ for $$\theta$$ in the interval $$0^\circ \leq \theta \leq 360^\circ$$. 2. **Isolate $$\sin \theta$$:** Subtract 3 from both sides: $$3 + 5 \sin \theta = 1 \implies 5 \sin \theta = 1 - 3$$ $$5 \sin \theta = -2$$ 3. **Divide both sides by 5:** $$\sin \theta = \frac{-2}{5}$$ Intermediate step showing cancellation: $$\sin \theta = \frac{\cancel{5} \times (-2/\cancel{5})}{\cancel{5}} = -\frac{2}{5}$$ 4. **Find the reference angle:** Calculate $$\alpha = \arcsin\left(\left| -\frac{2}{5} \right|\right) = \arcsin\left(\frac{2}{5}\right)$$. Using a calculator: $$\alpha \approx 23.58^\circ$$ 5. **Determine the solutions in $$0^\circ \leq \theta \leq 360^\circ$$:** Since $$\sin \theta$$ is negative, $$\theta$$ lies in the third and fourth quadrants. - Third quadrant: $$\theta = 180^\circ + \alpha = 180^\circ + 23.58^\circ = 203.58^\circ$$ - Fourth quadrant: $$\theta = 360^\circ - \alpha = 360^\circ - 23.58^\circ = 336.42^\circ$$ 6. **Final answer:** $$\theta \approx 203.6^\circ \text{ and } 336.4^\circ$$ These correspond to the choice: **203.6° and 336.4°**.