7. Let $\theta \in[0,2 \pi]$, if for any $x \in[0,1]$ there is always $2 x^{2} \sin \theta-4 x(1-x) \cos \theta+3(1-x)^{2}>0$, then the range of $\theta$ is $\qquad$ .
When $x=1$, we have $\sin \theta>0$, and $2 x^{2} \sin \theta+3(1-x)^{2} \geqslant 2 \sqrt{6 \sin \theta} x(1-x)$. When equality holds, $2 x^{2} \sin \theta=3(1-x)^{2}$. Therefore, we only need to consider $\sqrt{6 \sin \theta}-2 \cos \theta>0$.
From $\left\{\begin{array}{l}\cos \theta \geqslant 0, \\ 3 \sin \theta>2 \cos ^{2} \theta\end{array} \Rightarrow 2 \sin ^{2} \theta+3 \sin \theta-2>0 \Rightarrow \sin \theta>\frac{1}{2}\right.$ or $\cos \theta<0$, we know $\theta \in\left(\frac{\pi}{6}, \pi\right)$.
It is easy to verify that the original inequality holds in this case. Therefore, the range of $\theta$ is $\left(\frac{\pi}{6}, \pi\right)$.