Find, with proof, all real numbers $ x \in \lbrack 0, \frac {\pi}{2} \rbrack$, such that $ (2 \minus{} \sin 2x)\sin (x \plus{} \frac {\pi}{4}) \equal{} 1$.
1. We start with the given equation:
   \[
   (2 - \sin 2x) \sin \left( x + \frac{\pi}{4} \right) = 1
   \]

2. Using the double-angle identity for sine, we have:
   \[
   \sin 2x = 2 \sin x \cos x
   \]
   Substituting this into the equation, we get:
   \[
   (2 - 2 \sin x \cos x) \sin \left( x + \frac{\pi}{4} \right) = 1
   \]

3. Next, we use the angle addition formula for sine:
   \[
   \sin \left( x + \frac{\pi}{4} \right) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}
   \]
   Since \(\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}\), we have:
   \[
   \sin \left( x + \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} (\sin x + \cos x)
   \]

4. Substituting this back into the equation, we get:
   \[
   (2 - 2 \sin x \cos x) \cdot \frac{1}{\sqrt{2}} (\sin x + \cos x) = 1
   \]
   Simplifying, we have:
   \[
   \frac{2 - 2 \sin x \cos x}{\sqrt{2}} (\sin x + \cos x) = 1
   \]
   \[
   (2 - 2 \sin x \cos x) (\sin x + \cos x) = \sqrt{2}
   \]

5. Let \( l = \sin x + \cos x \) and \( m = \sin x \cos x \). Then the equation becomes:
   \[
   (2 - 2m) l = \sqrt{2}
   \]
   \[
   2l - 2ml = \sqrt{2}
   \]
   \[
   l(2 - 2m) = \sqrt{2}
   \]
   \[
   l = \frac{\sqrt{2}}{2 - 2m}
   \]

6. We also know from the Pythagorean identity:
   \[
   (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x
   \]
   \[
   l^2 = 1 + 2m
   \]

7. Substituting \( l = \frac{\sqrt{2}}{2 - 2m} \) into \( l^2 = 1 + 2m \), we get:
   \[
   \left( \frac{\sqrt{2}}{2 - 2m} \right)^2 = 1 + 2m
   \]
   \[
   \frac{2}{(2 - 2m)^2} = 1 + 2m
   \]
   \[
   2 = (1 + 2m)(2 - 2m)^2
   \]

8. Solving this cubic equation, we get:
   \[
   2 = (1 + 2m)(4 - 8m + 4m^2)
   \]
   \[
   2 = 4 + 8m^2 - 8m + 4m - 8m^2 + 2m
   \]
   \[
   2 = 4 - 4m + 4m^2
   \]
   \[
   4m^3 - 6m^2 + 1 = 0
   \]

9. The roots of the cubic equation \( 4m^3 - 6m^2 + 1 = 0 \) are \( m = \frac{1}{2}, 1 - \frac{\sqrt{3}}{2}, 1 + \frac{\sqrt{3}}{2} \).

10. Since \( \sin x \cos x \leq \frac{1}{2} \), the only feasible root is \( m = \frac{1}{2} \).

11. Substituting \( m = \frac{1}{2} \) back, we get:
    \[
    l^2 = 1 + 2 \cdot \frac{1}{2} = 2
    \]
    \[
    l = \sqrt{2}
    \]

12. Solving for \( \sin x \) and \( \cos x \) when \( \sin x + \cos x = \sqrt{2} \) and \( \sin x \cos x = \frac{1}{2} \), we get:
    \[
    \sin x = \cos x = \frac{1}{\sqrt{2}}
    \]
    \[
    x = \frac{\pi}{4}
    \]

The final answer is \( \boxed{ x = \frac{\pi}{4} } \)