4. Given the function
$$
f(x)=2^{2 x+1} \cos ^{2} \frac{\theta}{2}-2 \sqrt{2} \times 6^{x}+9^{x}(\sin \theta+1)
$$

is always positive on $\mathbf{R}$. Then the range of the real number $\theta$ is
$\qquad$
4. $\left(2 k \pi, 2 k \pi+\frac{\pi}{2}\right)(k \in \mathbf{Z})$.

The given condition is equivalent to the inequality
$$
4^{x}(\cos \theta+1)+9^{x}(\sin \theta+1)>2 \sqrt{2} \times 6^{x}
$$

which holds.
Dividing both sides of the above inequality by $6^{x}$, we get
$$
\left(\frac{2}{3}\right)^{x}(\cos \theta+1)+\left(\frac{3}{2}\right)^{x}(\sin \theta+1)>2 \sqrt{2} \text {. }
$$

Let $\left(\frac{2}{3}\right)^{x}=t$, then $t \in(0,+\infty)$. We now find the minimum value of
$$
\begin{array}{l}
\left(\frac{2}{3}\right)^{x}(\cos \theta+1)+\left(\frac{3}{2}\right)^{x}(\sin \theta+1) \\
=t(\cos \theta+1)+\frac{1}{t}(\sin \theta+1)
\end{array}
$$

By $t(\cos \theta+1)+\frac{1}{t}(\sin \theta+1)$
$$
\geqslant 2 \sqrt{(\cos \theta+1)(\sin \theta+1)},
$$

with equality holding if and only if $t^{2}=\frac{\sin \theta+1}{\cos \theta+1}$. Thus, the minimum value of $t(\cos \theta+1)+\frac{1}{t}(\sin \theta+1)$ is
$2 \sqrt{(\cos \theta+1)(\sin \theta+1)}$.
According to the problem, we have
$$
2 \sqrt{(\cos \theta+1)(\sin \theta+1)}>2 \sqrt{2} \text {, }
$$

which simplifies to $\sin \theta \cdot \cos \theta+\sin \theta+\cos \theta>1$.
Let $u=\sin \theta+\cos \theta=\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)$, then $u \in$ $[-\sqrt{2}, \sqrt{2}]$. Therefore, equation (1) becomes $u^{2}+2 u-3>0$.
Solving this, we get $u>1$.
Combining this with the range of $u$, we have $1<u \leqslant \sqrt{2}$.
Thus, $\frac{\sqrt{2}}{2}<\sin \left(\theta+\frac{\pi}{4}\right) \leqslant 1$.
Therefore, $2 k \pi+\frac{\pi}{4}<\theta+\frac{\pi}{4}<2 k \pi+\frac{3 \pi}{4}(k \in \mathbf{Z})$.
Hence, the range of $\theta$ is $\left(2 k \pi, 2 k \pi+\frac{\pi}{2}\right)(k \in \mathbf{Z})$.