1. **State the problem:** Solve for $\theta$ in degrees the equation $$2\sin^2(\theta) - 3\sin(\theta) + 1 = 0$$ where $0^\circ \leq \theta < 360^\circ$. 2. **Substitute:** Let $x = \sin(\theta)$, so the equation becomes $$2x^2 - 3x + 1 = 0$$. 3. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $a=2$, $b=-3$, $c=1$. Calculate the discriminant: $$\Delta = (-3)^2 - 4 \times 2 \times 1 = 9 - 8 = 1$$. So, $$x = \frac{3 \pm \sqrt{1}}{4} = \frac{3 \pm 1}{4}$$. 4. **Find roots:** - $$x_1 = \frac{3 + 1}{4} = 1$$ - $$x_2 = \frac{3 - 1}{4} = \frac{2}{4} = 0.5$$ 5. **Find $\theta$ for each root:** - For $x_1 = \sin(\theta) = 1$, $\theta = 90^\circ$ (since $\sin(90^\circ) = 1$). - For $x_2 = \sin(\theta) = 0.5$, $\theta$ can be $30^\circ$ or $150^\circ$ because $\sin(30^\circ) = \sin(150^\circ) = 0.5$. 6. **Check the domain:** All solutions $30^\circ$, $90^\circ$, and $150^\circ$ lie within $0^\circ \leq \theta < 360^\circ$. **Final answer:** $$\theta = 30^\circ, 90^\circ, 150^\circ$$.