1. **State the problem:** Solve the trigonometric equation $$\cos \alpha - 4 \sin^2 \alpha = 3$$ for $$0^\circ \leq \alpha \leq 360^\circ$$. 2. **Use the Pythagorean identity:** Recall that $$\sin^2 \alpha = 1 - \cos^2 \alpha$$. 3. **Substitute into the equation:** $$\cos \alpha - 4(1 - \cos^2 \alpha) = 3$$ 4. **Expand and simplify:** $$\cos \alpha - 4 + 4 \cos^2 \alpha = 3$$ 5. **Rearrange terms:** $$4 \cos^2 \alpha + \cos \alpha - 4 = 3$$ 6. **Bring all terms to one side:** $$4 \cos^2 \alpha + \cos \alpha - 7 = 0$$ 7. **Let $$x = \cos \alpha$$, then solve the quadratic:** $$4x^2 + x - 7 = 0$$ 8. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 4 \cdot (-7)}}{2 \cdot 4} = \frac{-1 \pm \sqrt{1 + 112}}{8} = \frac{-1 \pm \sqrt{113}}{8}$$ 9. **Calculate the roots:** $$x_1 = \frac{-1 + \sqrt{113}}{8} \approx \frac{-1 + 10.63}{8} = \frac{9.63}{8} = 1.20375$$ $$x_2 = \frac{-1 - \sqrt{113}}{8} \approx \frac{-1 - 10.63}{8} = \frac{-11.63}{8} = -1.45375$$ 10. **Check the domain of cosine:** Since $$\cos \alpha$$ must be between $$-1$$ and $$1$$, both values are invalid. 11. **Conclusion:** There are no real solutions for $$\alpha$$ in $$0^\circ \leq \alpha \leq 360^\circ$$ because the quadratic solutions for $$\cos \alpha$$ are outside the valid range. **Final answer:** No solutions in the given range.