3. Given $x \in \mathbf{C}, \arg \left(x^{2}-2\right)=\frac{3}{4} \pi, \arg \left(x^{2}\right.$ $+2 \sqrt{3})=\frac{\pi}{6}$. Then the value of $x$ is $\qquad$ .
3. $\pm(1+i)$.

Let $x^{2}-2=\rho_{1}\left(\cos \frac{3}{4} \pi+i \sin \frac{3}{4} \pi\right)$,
$$
x^{2}+2 \sqrt{3}=\rho_{2}\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right) \text {. }
$$
(2) - (1) gives
$2+2 \sqrt{3}=\left(\frac{\sqrt{3}}{2} \rho_{2}+\frac{\sqrt{2}}{2} \rho_{1}\right)+i\left(\frac{1}{2} \rho_{2}-\frac{\sqrt{2}}{2} \rho_{1}\right)$.
Thus, we have $\left\{\begin{array}{l}\frac{\sqrt{3}}{2} \rho_{2}+\frac{\sqrt{2}}{2} \rho_{1}=2+2 \sqrt{3} \\ \frac{1}{2} \rho_{2}-\frac{\sqrt{2}}{2} \rho_{1}=0 .\end{array}\right.$
Solving these equations, we get $\left\{\begin{array}{l}\rho_{1}=2 \sqrt{2}, \\ \rho_{2}=4 .\end{array}\right.$
Substituting into (1) gives $x^{2}=2i$.
Therefore, $x= \pm(1+i)$.