3. Given $3 \sqrt{2} \cos \left(\theta+\frac{\pi}{4}\right)<4\left(\sin ^{3} \theta-\cos ^{3} \theta\right)$, $\theta \in[-\pi, \pi]$. Then the range of $\theta$ is $\qquad$ .
3. $\left[-\pi,-\frac{3 \pi}{4}\right) \cup\left(\frac{\pi}{4}, \pi\right]$.

From the original inequality, we have
$$
\begin{array}{l}
3(\cos \theta-\sin \theta)<4\left(\sin ^{3} \theta-\cos ^{3} \theta\right) \\
\Leftrightarrow 4 \cos ^{3} \theta+3 \cos \theta<4 \sin ^{3} \theta+3 \sin \theta .
\end{array}
$$

Since $f(x)=4 x^{3}+3 x$ is a monotonically increasing function,
$$
\begin{array}{l}
\Rightarrow \cos \theta<\sin \theta \\
\Rightarrow x \in\left[-\pi,-\frac{3 \pi}{4}\right) \cup\left(\frac{\pi}{4}, \pi\right] .
\end{array}
$$