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(A) Given the determinant $f(x) = \left| \begin{array}{ccc} \sin^2 x & -2 + \cos^2 x & \cos 2x \ 2 + \sin^2 x & \cos^2 x & \cos 2x \ \sin^2 x & \cos

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$6$

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(A) Given the determinant $f(x) = \left| \begin{array}{ccc} \sin^2 x & -2 + \cos^2 x & \cos 2x \ 2 + \sin^2 x & \cos^2 x & \cos 2x \ \sin^2 x & \cos^2 x & 1 + \cos 2x \end{array} ight|$.Applying row operations $R_1 ightarrow R_1 - R_2$ and $R_2 ightarrow R_2 - R_3$:$f(x) = \left| \begin{array}{ccc} -2 & -2 & 0 \ 2 & 0 & -1 \ \sin^2 x & \cos^2 x & 1 + \cos 2x \end{array} ight|$.Expanding along the first row:$f(x) = -2(0 - (-\cos^2 x)) - (-2)(2(1 + \cos 2x) - (-\sin^2 x)) + 0$$f(x) = -2 \cos^2 x + 2(2 + 2 \cos 2x + \sin^2 x)$$f(x) = -2 \cos^2 x + 4 + 4 \cos 2x + 2 \sin^2 x$$f(x) = 4 + 4 \cos 2x - 2(\cos^2 x - \sin^2 x)$Since $\cos 2x = \cos^2 x - \sin^2 x$,we have:$f(x) = 4 + 4 \cos 2x - 2 \cos 2x = 4 + 2 \cos 2x$.For $x \in [0, \pi]$,the maximum value of $\cos 2x$ is $1$.Therefore,$f(x)_{	ext{max}} = 4 + 2(1) = 6$.

$f(x) = \left| \begin{array}{ccc} \sin^2 x & -2 + \cos^2 x & \cos 2x \\ 2 + \sin^2 x & \cos^2 x & \cos 2x \\ \sin^2 x & \cos^2 x & 1 + \cos 2x \end{array} \right|, x \in [0, \pi]$. The maximum value of $f(x)$ is equal to $.....$

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