3. Given a function $f(x)$ defined on the domain $[-1,1]$ that satisfies: for $\theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, 
$$
f(\sin \theta)+3 f(-\sin \theta)=\cos \theta \text {. }
$$

Then the range of the function $f(x)$ is
3. $\left[0, \frac{1}{4}\right]$.

In the given equation, replace $\theta$ with $-\theta$ to get
$$
f(-\sin \theta)+3 f(\sin \theta)=\cos \theta \text {. }
$$

Solving the two equations simultaneously, we get
$$
f(\sin \theta)=\frac{1}{4} \cos \theta=\frac{1}{4} \sqrt{1-\sin ^{2} \theta} \text {. }
$$

Therefore, $f(x)=\frac{1}{4} \sqrt{1-x^{2}}(x \in[-1,1])$.
Thus, the range of the function $f(x)$ is $\left[0, \frac{1}{4}\right]$.