1. **State the problem:** Simplify the expression $$\frac{1 - \sin x}{1 + \cos 2x}$$. 2. **Recall relevant formulas:** - Double angle formula for cosine: $$\cos 2x = 2\cos^2 x - 1$$. - Pythagorean identity: $$\sin^2 x + \cos^2 x = 1$$. 3. **Rewrite the denominator using the double angle formula:** $$1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x$$. 4. **Substitute back into the expression:** $$\frac{1 - \sin x}{2\cos^2 x}$$. 5. **Express numerator in terms of cosine if possible or simplify as is:** We can leave numerator as is for now. 6. **Rewrite denominator to show intermediate cancellation if needed:** $$\frac{1 - \sin x}{2\cos^2 x} = \frac{1 - \sin x}{2\cos x \cdot \cos x}$$. 7. **Multiply numerator and denominator by the conjugate of numerator to simplify:** Multiply numerator and denominator by $$1 + \sin x$$: $$\frac{(1 - \sin x)(1 + \sin x)}{2\cos^2 x (1 + \sin x)} = \frac{1 - \sin^2 x}{2\cos^2 x (1 + \sin x)}$$. 8. **Use Pythagorean identity in numerator:** $$1 - \sin^2 x = \cos^2 x$$. 9. **Substitute and simplify:** $$\frac{\cos^2 x}{2\cos^2 x (1 + \sin x)} = \frac{\cancel{\cos^2 x}}{2\cancel{\cos^2 x} (1 + \sin x)} = \frac{1}{2(1 + \sin x)}$$. **Final answer:** $$\boxed{\frac{1}{2(1 + \sin x)}}$$