3. The function
$$
f(x)=\left|\sin x+\frac{1}{2} \sin 2 x\right|(x \in \mathbf{R})
$$

has the range
3. $\left[0, \frac{3 \sqrt{3}}{4}\right]$.

Obviously, $f(0)=0, f(x) \geqslant 0$.
Therefore, the minimum value of $f(x)$ is 0.
Next, we find the maximum value of $f(x)$.
Notice that
$$
\begin{array}{l}
f(x)=\left|\sin x+\frac{1}{2} \sin 2 x\right| \\
=\left|(1+\cos x) \sin x \right| \\
=\left|2 \cos ^{2} \frac{x}{2} \cdot 2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}\right| \\
=4\left|\sin \frac{x}{2} \cdot \cos ^{3} \frac{x}{2}\right| \\
=4 \sqrt{27 \sin ^{2} \frac{x}{2} \cdot \frac{1}{3} \cos ^{2} \frac{x}{2} \cdot \frac{1}{3} \cos ^{2} \frac{x}{2} \cdot \frac{1}{3} \cos ^{2} \cdot \frac{x}{2}} \\
\leqslant 4 \sqrt{27\left(\frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}}{4}\right)^{4}}=\frac{3 \sqrt{3}}{4} .
\end{array}
$$

The equality holds if and only if $\sin ^{2} \frac{x}{2}=\frac{1}{3} \cos ^{2} \frac{x}{2}$, i.e., $x=2 k \pi \pm \frac{\pi}{3}(k \in \mathbf{Z})$.
Therefore, the maximum value of $f(x)$ is $\frac{3 \sqrt{3}}{4}$.
Thus, the range of $f(x)$ is $\left[0, \frac{3 \sqrt{3}}{4}\right]$.