1. **State the problem:** Solve the equation $$\sin^2\theta - \cos^2\theta - \cos\theta + 1 = 0$$ for $$0 \leq \theta < 2\pi$$. 2. **Use the Pythagorean identity:** Recall that $$\sin^2\theta = 1 - \cos^2\theta$$. 3. **Substitute into the equation:** $$ (1 - \cos^2\theta) - \cos^2\theta - \cos\theta + 1 = 0 $$ 4. **Simplify the equation:** $$ 1 - \cos^2\theta - \cos^2\theta - \cos\theta + 1 = 0 $$ $$ 2 - 2\cos^2\theta - \cos\theta = 0 $$ 5. **Rearrange terms:** $$ 2 - \cos\theta - 2\cos^2\theta = 0 $$ 6. **Multiply both sides by 1 (no change) and rewrite:** $$ -2\cos^2\theta - \cos\theta + 2 = 0 $$ 7. **Multiply entire equation by -1 to simplify:** $$ 2\cos^2\theta + \cos\theta - 2 = 0 $$ 8. **Let $$x = \cos\theta$$, then solve quadratic:** $$ 2x^2 + x - 2 = 0 $$ 9. **Use quadratic formula:** $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 16}}{4} = \frac{-1 \pm \sqrt{17}}{4} $$ 10. **Calculate roots:** $$ x_1 = \frac{-1 + \sqrt{17}}{4} \approx 0.7808, \quad x_2 = \frac{-1 - \sqrt{17}}{4} \approx -1.2808 $$ 11. **Check domain for cosine:** Since $$\cos\theta$$ must be between -1 and 1, $$x_2$$ is invalid. 12. **Find $$\theta$$ for valid root:** $$ \cos\theta = 0.7808 $$ 13. **Find $$\theta$$ in $$[0, 2\pi)$$:** $$ \theta = \pm \arccos(0.7808) + 2k\pi $$ Within $$0 \leq \theta < 2\pi$$, solutions are: $$ \theta_1 = \arccos(0.7808) \approx 0.6747 $$ $$ \theta_2 = 2\pi - 0.6747 = 5.6085 $$ **Final answer:** $$\theta \approx 0.6747, 5.6085$$ radians.