Example 6 Given the inequality $\sqrt{2}(2 a+3) \cos \left(\theta-\frac{\pi}{4}\right)+\frac{6}{\sin \theta+\cos \theta}-2 \sin 2 \theta<3 a+6$ holds for $\theta \in\left[0, \frac{\pi}{2}\right]$, find the range of values for $a$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Given that there are two variables in the inequality, and one of their ranges is provided, the method of separating variables is commonly adopted to find the range of the other. Noting the several trigonometric expressions related to angle $\theta$, $\cos \left(\theta-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}(\sin \theta + \cos \theta)$, and $\sin 2 \theta = 2 \sin \theta \cos \theta$, we consider letting $\sin \theta + \cos \theta = x$ for variable substitution to simplify the given inequality and then seek a solution.

Solution: Let $\sin \theta + \cos \theta = x$, then $\cos \left(\theta-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} x$, and $\sin 2 \theta = x^2 - 1$. When $\theta \in \left[0, \frac{\pi}{2}\right]$, $x \in [1, \sqrt{2}]$. Thus, the original inequality can be transformed into $(2a + 3)x + \frac{6}{x} - 2(x^2 - 1) = 0.2x\left(x + \frac{2}{x} - a\right) - 3\left(x + \frac{2}{x} - a\right) > 0$, which simplifies to $(2x - 3)\left(x + \frac{2}{x} - a\right) > 0$ for $x \in [1, \sqrt{2}]$.

Therefore, the original inequality is equivalent to inequality (1). Since $x \in [1, \sqrt{2}]$, we have $2x - 3 < 0$. Also, $\left(x + \frac{2}{x}\right)_{\max}(x \in [1, \sqrt{2}])$. Since $f(x) = x + \frac{2}{x}$ is decreasing on $[1, \sqrt{2}]$, $\left(x + \frac{2}{x}\right)_{\max} = 3$ for $x \in [1, \sqrt{2}]$. Therefore, $a > 3$.