3. Let the vectors be
$$
\begin{array}{l}
\boldsymbol{\alpha}=(x+3, x), \\
\boldsymbol{\beta}=(2 \sin \theta \cdot \cos \theta, a \sin \theta+a \cos \theta)
\end{array}
$$

satisfy that for any $x \in \mathbf{R}$ and $\theta \in\left[0, \frac{\pi}{2}\right]$, we have
$$
|\boldsymbol{\alpha}+\boldsymbol{\beta}| \geqslant \sqrt{2} \text {. }
$$

Then the range of the real number $a$ is $\qquad$
3. $a \leqslant 1$ or $a \geqslant 5$.

Let $\sin \theta + \cos \theta = t$. Then
$t \in [1, \sqrt{2}]$,
$$
\begin{array}{l}
2 \sin \theta \cdot \cos \theta = t^{2} - 1, \\
\boldsymbol{\alpha} + \boldsymbol{\beta} = (t^{2} + x + 2, x + a t).
\end{array}
$$

Notice that,
$$
\begin{array}{l}
\left(t^{2} + x + 2\right)^{2} + (x + a t)^{2} \\
\geqslant \frac{1}{2} \left(t^{2} + x + 2 - x - a t\right)^{2} \\
= \frac{1}{2} \left(t^{2} - a t + 2\right)^{2}.
\end{array}
$$

Then $|\boldsymbol{\alpha} + \boldsymbol{\beta}| \geqslant \sqrt{2}$
$\Leftrightarrow \left(t^{2} + x + 2\right)^{2} + (x + a t)^{2} \geqslant 2$ for any $x \in \mathbf{R}$
$\Leftrightarrow \frac{1}{2} \left(t^{2} - a t + 2\right)^{2} \geqslant 2$
$\Leftrightarrow t^{2} - a t \geqslant 0$ or $t^{2} - a t + 4 \leqslant 0$
$\Leftrightarrow a \leqslant t$ or $a \geqslant t + \frac{4}{t}$ for any $t \in [1, \sqrt{2}]$
$\Leftrightarrow a \leqslant 1$ or $a \geqslant 5$.