3. Given the complex number $z$ satisfies $\left|2 z+\frac{1}{z}\right|=1$, then the principal value range of the argument of $z$ is $\qquad$
3. $k \pi+\frac{\pi}{2}-\frac{1}{2} \arccos \frac{3}{4} \leqslant \theta \leqslant k \pi+\frac{\pi}{2}+\frac{1}{2} \arccos \frac{3}{4}(k=0,1)$.

Let $z=r(\cos \theta+\sin \theta)$, then
$$
\begin{array}{l}
\left|2 z+\frac{1}{z}\right|^{2}=\left(4 r^{2}+\frac{1}{r^{2}}\right)+4 \cos 2 \theta=1, \\
4 r^{4}+(4 \cos 2 \theta-1) r^{2}+1=0 .
\end{array}
$$

This equation holds if and only if the quadratic equation
$$
4 x^{2}+(4 \cos 2 \theta-1) x+1=0
$$

has two positive roots $x_{1}$ and $x_{2}$
$$
\begin{array}{l}
\Leftrightarrow \text { discriminant } \Delta=(4 \cos 2 \theta-1)^{2}-4^{2} \geqslant 0 \text { and } \\
x_{1}+x_{2}=-\frac{1}{4}(4 \cos 2 \theta-1)>0, x_{1} x_{2}=\frac{1}{4}>0 \\
\Leftrightarrow \cos 2 \theta \leqslant-\frac{3}{4} \\
\Leftrightarrow 2 k \pi+\pi-\arccos \frac{3}{4} \leqslant 2 \theta \leqslant 2 k \pi+\pi+\arccos \frac{3}{4} \\
\Leftrightarrow k \pi+\frac{\pi}{2}-\frac{1}{2} \arccos \frac{3}{4} \leqslant \theta \leqslant k \pi+\frac{\pi}{2}+\frac{1}{2} \arccos \frac{3}{4}(k=0,1) .
\end{array}
$$