1. **State the problem:** Solve the equation $$6 \cos^2 \theta - 4 = \sin \theta$$ for $$0^\circ \leq \theta \leq 360^\circ$$. 2. **Use the Pythagorean identity:** Recall that $$\cos^2 \theta = 1 - \sin^2 \theta$$. 3. **Substitute into the equation:** $$6(1 - \sin^2 \theta) - 4 = \sin \theta$$ 4. **Simplify:** $$6 - 6 \sin^2 \theta - 4 = \sin \theta$$ $$2 - 6 \sin^2 \theta = \sin \theta$$ 5. **Rearrange to form a quadratic in $$\sin \theta$$:** $$-6 \sin^2 \theta - \sin \theta + 2 = 0$$ 6. **Multiply both sides by $$-1$$ to simplify:** $$\cancel{-6} \sin^2 \theta + \cancel{-} \sin \theta - \cancel{2} = 0 \Rightarrow 6 \sin^2 \theta + \sin \theta - 2 = 0$$ 7. **Solve the quadratic equation:** Use the quadratic formula $$\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=6$$, $$b=1$$, $$c=-2$$. Calculate the discriminant: $$\Delta = 1^2 - 4 \times 6 \times (-2) = 1 + 48 = 49$$ Calculate roots: $$\sin \theta = \frac{-1 \pm \sqrt{49}}{2 \times 6} = \frac{-1 \pm 7}{12}$$ 8. **Find the two possible values:** - $$\sin \theta = \frac{-1 + 7}{12} = \frac{6}{12} = 0.5$$ - $$\sin \theta = \frac{-1 - 7}{12} = \frac{-8}{12} = -\frac{2}{3} \approx -0.6667$$ 9. **Find $$\theta$$ for each value in the interval $$0^\circ \leq \theta \leq 360^\circ$$:** - For $$\sin \theta = 0.5$$, $$\theta = 30^\circ$$ or $$150^\circ$$. - For $$\sin \theta = -\frac{2}{3}$$, $$\theta = 360^\circ - \arcsin\left(\frac{2}{3}\right) \approx 360^\circ - 41.81^\circ = 318.19^\circ$$ and $$\theta = 180^\circ + 41.81^\circ = 221.81^\circ$$. 10. **Final solutions:** $$\theta = 30^\circ, 150^\circ, 221.81^\circ, 318.19^\circ$$