6

Let

$$a=\sin x+\cos x,b=\sin x\cos x$$

We know that ${\sin }^{2}x+{\cos }^{2}x=1$, so we have

$$(\sin x+\cos x{)}^2={\sin }^{2}x+2\sin x\cos x+{\cos }^{2}x=1+2b$$

So

$$a^2=1+2b⟹b=\frac{a^2-1}{2}$$

We have the equation

$$2(\sin x+\cos x)+(\sec x+\csc x)=3\sqrt{6}$$

It can be re-write as

$$\begin{array}{rl}2a+(\sec x+\csc x)& =3\sqrt{6}\\ 2a+\frac{1}{\cos x}+\frac{1}{\sin x}& =3\sqrt{6}\\ 2a+\frac{\sin x+\cos x}{\sin x\cos x}& =3\sqrt{6}\\ 2a+\frac{a}{b}& =3\sqrt{6}\end{array}$$

Also we have

$$\frac{a}{b}=\frac{a}{\frac{b^2-1}{2}}=\frac{2a}{a^2-1}$$

Then the equation is

$$\begin{array}{rl}2a+\frac{2a}{a^2-1}& =3\sqrt{6}\\ \frac{2a^3-2a+2a}{a^2-1}& =3\sqrt{6}\\ \frac{2a^3}{a^2-1}& =3\sqrt{6}\\ 2a^3& =3\sqrt{6}(a^2-1)\\ 2a^3-3\sqrt{6}{a}^2+3\sqrt{6}& =0\end{array}$$

Let

$$a=c+\frac{\sqrt{6}}{2}$$

Then we know

$$a^3=(c+\frac{\sqrt{6}}{2}{)}^3=c^3+3c^2\frac{\sqrt{6}}{2}+3c(\frac{\sqrt{6}}{2}{)}^2+(\frac{\sqrt{6}}{2}{)}^3=c^3+\frac{3\sqrt{6}}{2}{c}^2+\frac{9}{2}c+\frac{3\sqrt{6}}{4}$$ $$a^2=(c+\frac{\sqrt{6}}{2}{)}^2=c^2+2c\frac{\sqrt{6}}{2}+(\frac{\sqrt{6}}{2}{)}^2=c^2+\sqrt{6}c+\frac{3}{2}$$

Substitute them into the function

$$\begin{array}{rl}2(c^3+\frac{3\sqrt{6}}{2}{c}^2+\frac{9}{2}c+\frac{3\sqrt{6}}{4})-3\sqrt{6}(c^2+\sqrt{6}c+\frac{3}{2})+3\sqrt{6}& =0\\ 2c^3+3\sqrt{6}{c}^2+9c+\frac{3\sqrt{6}}{2}-3\sqrt{6}{c}^2-18c-\frac{9\sqrt{6}}{2}+3\sqrt{6}& =0\\ 2c^3-9c& =0\end{array}$$

We can get the 3 roots

$$\{\begin{array}{rl}& c_1=0\\ & c_2=\frac{3\sqrt{2}}{2}\\ & c_3=-\frac{3\sqrt{2}}{2}\end{array}$$

Back to $a$, we have

$$\{\begin{array}{rl}& a_1=0+\frac{\sqrt{6}}{2}=\frac{\sqrt{6}}{2}\\ & a_2=\frac{3\sqrt{2}}{2}+\frac{\sqrt{6}}{2}=\frac{3\sqrt{2}+\sqrt{6}}{2}\\ & a_3=-\frac{3\sqrt{2}}{2}+\frac{\sqrt{6}}{2}=\frac{\sqrt{6}-3\sqrt{2}}{2}\end{array}$$

For any $x$, $-\sqrt{2}\le \sin x+\cos x\le \sqrt{2}$. So only $a=\frac{\sqrt{6}}{2}$ and $a=\frac{\sqrt{6}-3\sqrt{2}}{2}$ match the condition
Condition 1: $\sin x+\cos x=\frac{\sqrt{6}}{2}$, we have

$$\sin x+\cos x=\sqrt{2}\sin (x+\frac{\pi }{4})=\frac{\sqrt{6}}{2}⟹\sin (x+\frac{\pi }{4})=\frac{\sqrt{3}}{2}$$

So, for $k\in \mathbb{Z}$, we get

$$x+\frac{\pi }{4}=\frac{\pi }{3}+2k\pi \text{or}x+\frac{\pi }{4}=\frac{2\pi }{3}+2k\pi$$

and

$$\{\begin{array}{rl}& x=\frac{\pi }{12}+2k\pi \\ & x=\frac{5\pi }{12}+2k\pi \end{array}$$

Condition 2: $\sin x+\cos x=\frac{\sqrt{6}-3\sqrt{2}}{2}$, we have

Let $\theta =\arcsin \left(\frac{\sqrt{3}-3}{2}\right)$
So, for $k\in \mathbb{Z}$, we get

and

$$\{\begin{array}{rl}& x=\theta -\frac{\pi }{4}+2k\pi \\ & x=\frac{3\pi }{4}-\theta +2k\pi \end{array}$$

So we get the final answer

$$\boxed{k\in \mathbb{Z}\{\begin{array}{rl}& x=\frac{\pi }{12}+2k\pi \\ & x=\frac{5\pi }{12}+2k\pi \\ & x=\arcsin (\frac{\sqrt{3}-3}{2})-\frac{\pi }{4}+2k\pi \\ & x=\frac{3\pi }{4}-\arcsin (\frac{\sqrt{3}-3}{2})+2k\pi \end{array}}$$