3. The acute angle $x=$ $\qquad$ (in radians) that satisfies the equation $(\sin 2 x+\cos x)(\sin x-\cos x)=\cos x$.
(Provided by An Zhenping)
3. $\frac{\pi}{3}$.

The given condition can be transformed into
$$
(2 \sin x+1)(\sin x-\cos x)=1 \text {. }
$$

Since when $00$, at this time equation (1) does not hold. Therefore, it can only be that $\frac{\pi}{4}0$. So, $f(x)=(2 \sin x+1)(\sin x-\cos x)$ is an increasing function on $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, and
$$
f\left(\frac{\pi}{3}\right)=(\sqrt{3}+1)\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)=1,
$$

That is to say, the original equation can be transformed into $f(x)=f\left(\frac{\pi}{3}\right)$, so $x=\frac{\pi}{3}$.