1. **State the problem:** Simplify and solve the equation $$\frac{\sin\left(\frac{\pi}{2} - x\right) - 1}{1 - \cos(-x)} = \cos^2 x$$. 2. **Recall trigonometric identities:** - $$\sin\left(\frac{\pi}{2} - x\right) = \cos x$$. - $$\cos(-x) = \cos x$$ (cosine is an even function). 3. **Substitute identities into the equation:** $$\frac{\cos x - 1}{1 - \cos x} = \cos^2 x$$. 4. **Simplify the fraction:** Note that $$1 - \cos x = -(\cos x - 1)$$, so $$\frac{\cos x - 1}{1 - \cos x} = \frac{\cos x - 1}{-(\cos x - 1)} = -1$$ (for $$\cos x \neq 1$$). 5. **Rewrite the equation:** $$-1 = \cos^2 x$$. 6. **Analyze the equation:** Since $$\cos^2 x \geq 0$$ for all real $$x$$, the equation $$-1 = \cos^2 x$$ has no real solutions. 7. **Check the excluded case $$\cos x = 1$$:** If $$\cos x = 1$$, then denominator $$1 - \cos x = 0$$, which is undefined, so no solution there. **Final answer:** There are no real values of $$x$$ satisfying the given equation.