5. Given the inequality $\sqrt{2}(2 a+3) \cos \left(\theta-\frac{\pi}{4}\right)+\frac{6}{\sin \theta+\cos \theta}-2 \sin 2 \theta<3 a+6$, for $\theta \in\left[0, \frac{\pi}{2}\right]$ to always hold. Find the range of $\theta$. (1st China Southeast Mathematical Olympiad)
5. Let $\sin \theta+\cos \theta=x$, then $\cos \left(\theta-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} x, \sin 2 x=x^{2}-1, x \in[1, \sqrt{2}]$, thus the original inequality can be transformed into $(2 a+3) x+\frac{6}{x}-2\left(x^{2}-1\right)0$

Rearranging, we get
$$(2 x-3)\left(x+\frac{2}{x}-a\right)>0, x \in[1, \sqrt{2}]$$

Since $x \in[1, \sqrt{2}]$, we have $2 x-3\left(x+\frac{2}{x}\right)_{\max }(x \in[1, \sqrt{2}])$

It is easy to see that $f(x)=x+\frac{2}{x}$ is a decreasing function on $[1, \sqrt{2}]$, so $\left(x+\frac{2}{x}\right)_{\text {max }}=3,(x \in[1, \sqrt{2}])$. Therefore, $a>3$: