Let $$f(x) = \begin{vmatrix} \sin^2 x & -2 + \cos^2 x & \cos 2x \\ 2 + \sin^2 x & \cos^2 x & \cos 2x \\ \sin^2 x & \cos^2 x & 1 + \cos 2x \end{vmatrix}$$, $$x \in [0, \pi]$$. Then the maximum value of $$f(x)$$ is equal to _________.

We begin with the determinant-valued function

$$f(x)=\begin{vmatrix} \sin^{2}x & -2+\cos^{2}x & \cos 2x\\[4pt] 2+\sin^{2}x & \cos^{2}x & \cos 2x\\[4pt] \sin^{2}x & \cos^{2}x & 1+\cos 2x \end{vmatrix},\qquad x\in[0,\pi].$$

For compactness let us put

$$s=\sin^{2}x,\qquad c=\cos^{2}x.$$

Using the well-known double-angle identity $$\cos 2x=\cos^{2}x-\sin^{2}x=c-s,$$ we can rewrite every entry of the matrix in terms of these symbols.

The third-row, third-column entry becomes

$$1+\cos 2x = 1 + (c-s)=1+c-s.$$

Consequently the matrix takes the form

$$ \begin{vmatrix} s & -2+c & c-s\\[4pt] 2+s & c & c-s\\[4pt] s & c & 1+c-s \end{vmatrix}. $$

To simplify the determinant, we apply elementary row operations, remembering that adding or subtracting a multiple of one row from another does not change the value of the determinant.

First subtract the first row from the second:

$$R_2 \longleftarrow R_2-R_1,$$

which produces

$$R_2:\;(2+s)-s=2$$, $$\quad c-(-2+c)=2$$, $$\quad (c-s)-(c-s)=0.$$

$$ \begin{vmatrix} s & -2+c & c-s\\[4pt] 2 & 2 & 0\\[4pt] s & c & 1+c-s \end{vmatrix}. $$

Next, subtract the first row from the third:

$$R_3 \longleftarrow R_3-R_1,$$

to obtain

$$R_3:\;s-s=0$$, $$\quad c-(-2+c)=2$$, $$\quad (1+c-s)-(c-s)=1.$$

After these two operations the matrix becomes

$$ \begin{vmatrix} s & -2+c & c-s\\[4pt] 2 & 2 & 0\\[4pt] 0 & 2 & 1 \end{vmatrix}. $$

Now we will operate $$C_2 \longleftarrow C_2-C_1,$$

$$ \begin{vmatrix} s & -2+c-S & c-s\\[4pt] 2 & 2-2 & 0\\[4pt] 0 & 2-0 & 1 \end{vmatrix}. $$

$$\therefore$$ we get  $$\begin{vmatrix} s & -2+c-S & c-s\\[4pt] 2 & 0 & 0\\[4pt] 0 & 2 & 1 \end{vmatrix}. $$

Now we expand the determinant along the second row, using the cofactor expansion rule “$$+\;-\;+$$”:

$$ \begin{aligned} f(x) &= -2 \;\begin{vmatrix}2+c-s & c-s\\ 2 & 1\end{vmatrix} \; \end{aligned} $$

Now we expand it we get $$f(x)=-2\left(\left(-2+c-s\right)-2\left(c-s\right)\right)$$

$$\Rightarrow$$ $$f(x)=4-2s+2c$$

Recall that $$s-c=1-2c$$ 

Therefore, $$ f(x)=2+4c $$

The cosine function satisfies $$0\le\cos ^2x\le 1$$ for every real $$x$$, and this remains true on the interval $$[0,\pi]$$. Consequently

$$ 2+4(0)\le f(x)\le 2+4(1)\;\Longrightarrow\;2\le f(x)\le 6. $$

Hence the maximum value attained by $$f(x)$$ is

$$ f_{\max}=4+2\cdot1=6. $$

So, the answer is $$6$$.