1. **State the problem:** Solve for $x$ in the equation $$4 \sin^2 x - 4 \sin x + 1 = 0$$ where $$0^\circ \leq x \leq 360^\circ$$. 2. **Identify the substitution:** Let $$y = \sin x$$. The equation becomes $$4y^2 - 4y + 1 = 0$$. 3. **Solve the quadratic equation:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=4$$, $$b=-4$$, and $$c=1$$. 4. Calculate the discriminant: $$\Delta = (-4)^2 - 4 \times 4 \times 1 = 16 - 16 = 0$$. 5. Since $$\Delta = 0$$, there is one repeated root: $$y = \frac{-(-4)}{2 \times 4} = \frac{4}{8} = \frac{1}{2}$$. 6. **Back-substitute:** $$\sin x = \frac{1}{2}$$. 7. **Find all solutions for $$x$$ in $$0^\circ \leq x \leq 360^\circ$$:** $$x = 30^\circ, 150^\circ$$ because $$\sin 30^\circ = \sin 150^\circ = \frac{1}{2}$$. **Final answer:** $$x = 30^\circ, 150^\circ$$.