3. Find all real numbers $x \in\left[0, \frac{\pi}{2}\right]$, such that
$$
(2-\sin 2 x) \sin \left(x+\frac{\pi}{4}\right)=1 \text {, }
$$

and prove your conclusion.
(Li Shenghong)
3. Let $\sin \left(x+\frac{\pi}{4}\right)=t$, i.e., $\sin x+\cos x=\sqrt{2} t$. Then,

$1+\sin 2 x=2 t^{2}$, i.e., $\sin 2 x=2 t^{2}-1$. Therefore, $t\left(3-2 t^{2}\right)=1$,

which simplifies to $2 t^{3}-3 t+1=0$.
Notice that $t=1$ is a solution to the above equation, hence $(t-1)\left(2 t^{2}+2 t-1\right)=0$.
Since $0 \leqslant x \leqslant \frac{\pi}{2}$, it follows that $\frac{\sqrt{2}}{2} \leqslant t \leqslant 1$.
Thus, $2 t^{2}+2 t-1 \geqslant 2 \times \frac{1}{2}+2 \times \frac{\sqrt{2}}{2}-1>1$.
Therefore, the equation (1) has a unique solution $t=1$.
Hence, the original equation has a unique solution $x=\frac{\pi}{4}$.