3. Let the curve $f(x)=a \cos x+b \sin x$ have a line of symmetry at $x=\frac{\pi}{5}$. Then a point of symmetry for the curve $y=f\left(\frac{\pi}{10}-x\right)$ is ( ).
(A) $\left(\frac{\pi}{5}, 0\right)$
(B) $\left(\frac{2 \pi}{5}, 0\right)$
(C) $\left(\frac{3 \pi}{5}, 0\right)$
(D) $\left(\frac{4 \pi}{5}, 0\right)$
3. B.

Since $f(x)=a \cos x+b \sin x$
$$
=\sqrt{a^{2}+b^{2}} \sin (x+\theta)
$$

has a period of $2 \pi$, therefore, a point of symmetry for the curve
$$
f(x)=a \cos x+b \sin x
$$

is $\left(\frac{\pi}{5}+\frac{\pi}{2}, 0\right)$, which is $\left(\frac{7 \pi}{10}, 0\right)$.
Thus, a point of symmetry for the curve $y=f\left(x+\frac{\pi}{10}\right)$ is $\left(\frac{3 \pi}{5}, 0\right)$.

Then, a point of symmetry for the curve $y=f\left(\frac{\pi}{10}-x\right)$ is $\left(-\frac{3 \pi}{5}, 0\right)$.

Since the period of $y=f\left(\frac{\pi}{10}-x\right)$ is $2 \pi$, and the period of its points of symmetry is $\pi$, the answer is (B).