11. (2001 China Western Mathematical Olympiad) Find all real numbers $x \in\left[0, \frac{\pi}{2}\right]$, such that $(2-\sin 2 x)$ $\sin \left(x+\frac{\pi}{4}\right)=1$, and prove your conclusion.
11. Let $\sin \left(x+\frac{\pi}{4}\right)=t$, i.e., $\sin x + \cos x = \sqrt{2} t$. Therefore, $1 + \sin 2x = 2t^2$, which means $\sin 2x = 2t^2 - 1$. Thus, we have $t(3 - 2t^2) = 1$,
i.e., $2t^3 - 3t + 1 = 0$.
Notice that $t = 1$ is a solution to the above equation, hence
$$
(t-1)(2t^2 + 2t - 1) = 0.
$$

Since $0 \leq x \leq \frac{\pi}{2}$, we have $\frac{\sqrt{2}}{2} \leq t \leq 1$.
Therefore, $2t^2 + 2t - 1 \geq 2 \times \frac{1}{2} + 2 \times \frac{\sqrt{2}}{2} - 1 > 1$.
Thus, the equation (1) has a unique solution $t = 1$.
Hence, the original equation has a unique solution $x = \frac{\pi}{4}$.