7. Let $\theta \in[0,2 \pi]$ satisfy that for $x \in[0,1]$,
$$
f(x)=2 x^{2} \sin \theta-4 x(1-x) \cos \theta+3(1-x)^{2}>0 \text {. }
$$

Then the range of $\theta$ is $\qquad$
7. $\left(\frac{\pi}{6}, \pi\right)$.

Notice that, $2 \sin \theta=f(1)>0$.
Let $x \in(0,1)$ satisfy
$2 x^{2} \sin \theta=3(1-x)^{2}$.
Then $f(x)=2 x(1-x)(\sqrt{6 \sin \theta}-2 \cos \theta)>0$.
Since $\sin \theta>0$, and $\sqrt{6 \sin \theta}-2 \cos \theta>0$, it follows that for $x \in[0,1]$, we have $f(x)>0$.

Given that $g(\theta)=\sqrt{6 \sin \theta}-2 \cos \theta$ is monotonically increasing on the interval $\left[0, \frac{\pi}{2}\right]$, and $g\left(\frac{\pi}{6}\right)=0$, thus, the range of $\theta$ is $\left(\frac{\pi}{6}, \pi\right)$.