Let $\mathbf{A} = \{x \in (0, \pi) - \{\frac{\pi}{2}\} : \log_{(2 / \pi)} |\sin x| + \log_{(2 / \pi)} |\cos x| = 2\}$ and $\mathbf{B} = \{x \geqslant 0 : \sqrt{x} (\sqrt{x} - 4) - 3 |\sqrt{x} - 2| + 6 = 0\}$. Then $n(\mathbf{A} \cup \mathbf{B})$ is equal to:

Step 1: Solve for set A

We start by solving the equation for set A. Using the logarithm property $\log_b M + \log_b N = \log_b (MN)$, we can combine the two logarithmic terms. The base of the logarithm is $2/\pi$, which is between 0 and 1.

Step 2: Simplify and solve for sin(2x)

Applying the logarithm property, we get $\log_{(2/\pi)} |\sin x \cos x| = 2$. Converting this to exponential form gives $|\sin x \cos x| = (2/\pi)^2$. We know that $\sin(2x) = 2 \sin x \cos x$, so $|\sin x \cos x| = \frac{1}{2} |\sin(2x)|$. Substituting this, we find $|\sin(2x)| = \frac{8}{\pi^2}$. Since $\pi \approx 3.14$, $\pi^2 \approx 9.86$, so $8/\pi^2 \approx 0.81$, which is less than 1.

Step 3: Find solutions for x in set A

Let $\sin \alpha = 8/\pi^2$ where $\alpha \in (0, \pi/2)$. Since $x \in (0, \pi) - \{\pi/2\}$, we have $2x \in (0, 2\pi) - \{\pi\}$. The equation $|\sin(2x)| = 8/\pi^2$ means $\sin(2x) = 8/\pi^2$ or $\sin(2x) = -8/\pi^2$. This gives four distinct solutions for $2x$ in $(0, 2\pi)$, which are $\alpha$, $\pi - \alpha$, $\pi + \alpha$, and $2\pi - \alpha$. None of these values are $\pi$. Therefore, there are 4 distinct values for $x$ in set A.

Step 4: Solve for set B (Case 1: $\sqrt{x} - 2 \ge 0$)

Now we solve the equation for set B. Let $y = \sqrt{x}$. The equation becomes $y(y-4) - 3|y-2| + 6 = 0$. We consider two cases for $|y-2|$. Case 1: $y - 2 \ge 0 \implies y \ge 2$. The equation becomes $y^2 - 4y - 3(y-2) + 6 = 0$, which simplifies to $y^2 - 7y + 12 = 0$.

Step 5: Find solutions for y in Case 1

Factoring the quadratic equation $y^2 - 7y + 12 = 0$, we get $(y-3)(y-4) = 0$. This gives solutions $y=3$ or $y=4$. Both solutions satisfy the condition $y \ge 2$.

Step 6: Solve for x in Case 1

Since $y = \sqrt{x}$, we have $\sqrt{x} = 3 \implies x = 9$ and $\sqrt{x} = 4 \implies x = 16$. Both $x=9$ and $x=16$ are valid solutions for set B.

Step 7: Solve for set B (Case 2: $\sqrt{x} - 2 < 0$)

Case 2: $y - 2 < 0 \implies 0 \le y < 2$. The equation becomes $y^2 - 4y - 3(-(y-2)) + 6 = 0$, which simplifies to $y^2 - 4y + 3y - 6 + 6 = 0$, or $y^2 - y = 0$.

Step 8: Find solutions for y in Case 2 and final count for B

Factoring $y^2 - y = 0$, we get $y(y-1) = 0$. This gives solutions $y=0$ or $y=1$. Both solutions satisfy the condition $0 \le y < 2$. Therefore, $\sqrt{x} = 0 \implies x = 0$ and $\sqrt{x} = 1 \implies x = 1$. So, set B has elements $\{0, 1, 9, 16\}$. Thus, $n(\mathbf{B}) = 4$.

Step 9: Calculate $n(\mathbf{A} \cup \mathbf{B})$

We found $n(\mathbf{A}) = 4$ and $n(\mathbf{B}) = 4$. Now we need to check for common elements. The elements of A are of the form $x = \frac{1}{2} \alpha$, $\frac{1}{2}(\pi - \alpha)$, $\frac{1}{2}(\pi + \alpha)$, $\frac{1}{2}(2\pi - \alpha)$, where $\alpha = \arcsin(8/\pi^2)$. Since $8/\pi^2 \approx 0.81$, $\alpha \approx 0.94$ radians. The values for $x$ in A are approximately $0.47, 1.10, 2.08, 2.65$. None of these are integers. The elements of B are $\{0, 1, 9, 16\}$, which are all integers. Therefore, $\mathbf{A} \cap \mathbf{B} = \emptyset$, so $n(\mathbf{A} \cap \mathbf{B}) = 0$. Thus, $n(\mathbf{A} \cup \mathbf{B}) = 4 + 4 - 0 = 8$.

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