The maximum value of the determinant of the matrix $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1 \end{bmatrix}$ is:

(where $x$ is real)

The determinant of the matrix $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1 \end{bmatrix}$ can be found using first row expansion:

$\det = 1 \cdot \begin{vmatrix} 1+\sin x & 1 \\ 1 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ 1+\cos x & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1+\sin x \\ 1+\cos x & 1 \end{vmatrix}$

Calculating each 2×2 determinant:

First: $(1+\sin x)(1) - (1)(1) = \sin x$

Second: $(1)(1) - (1)(1+\cos x) = -\cos x$

Third: $(1)(1) - (1+\sin x)(1+\cos x)$

$= 1 - (1 + \cos x + \sin x + \sin x \cos x)$

$= -\cos x - \sin x - \sin x \cos x$

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Combining the results:

$\det = 1(\sin x) - 1(-\cos x) + 1(-\cos x - \sin x - \sin x \cos x)$

$= \sin x + \cos x - \cos x - \sin x - \sin x \cos x$

$= -\sin x \cos x$

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Using the double angle formula $\sin x \cos x = \frac{1}{2}\sin 2x$:

$-\sin x \cos x = -\frac{1}{2}\sin 2x$

The maximum value of $\sin 2x$ is $1$, and the minimum value is $-1$.

The maximum value of $-\sin 2x$ is $-(-1) = 1$.

Therefore:

Maximum value $= -\frac{1}{2} \times (-1)$

$= \frac{1}{2}$

This occurs when $\sin 2x = -1$, at $x = -\frac{\pi}{4}, \frac{3\pi}{4}$, etc.

The maximum value of the determinant is $\frac{1}{2}$.