5 L know the inequality $\sqrt{2}(2 a+3) \cos \left(\theta-\frac{\pi}{4}\right) \div \frac{6}{\sin \theta+\cos \theta}$ $2 \sin 2 \theta-3 a+6$ holds for $\mathrm{f} \theta \in\left[0, \frac{\pi}{2}\right.$, find the range of values for $a$. (Zhang Juncheng provided the problem)
Let $\sin \theta + \cos \theta = x$, then
$$
\cos \left(\theta - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} x, \quad \sin 2\theta - x^2 - 1, \quad x \in [1, \sqrt{2}] \cdot 2 \vdots
$$

Thus, the original inequality is transformed into
$$
(2a + 3)x + \frac{6}{x} - 2(x^2 - 1) = 0. \\
2x\left(x + \frac{2}{x} - a\right) - 3\left(x + \frac{2}{x} - a\right) > 0. \\
(2x - 3)\left(x + \frac{2}{x} - a\right) > 0 \quad (x \in [1, \sqrt{2}]) .
\end{array}
$$

Therefore, the original inequality is equivalent to inequality (1).
Since $x \in [1, \sqrt{2}]$, we have $2x - 3 < 0$ and $\left(x + \frac{2}{x}\right)_{\max}(x \in [1, \sqrt{2}])$.

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

Therefore, $a > 3$.