When solving for the union of two sets, simplify the equations for each set, find the solutions, and count the total number of unique elements in the union.
We first solve for the set \( A \) by simplifying the given equation and finding the range of \( x \) that satisfies it. Next, we solve for the set \( B \) using the given equation. After determining the elements in both sets, we calculate \( n(A \cup B) \), the number of elements in the union of sets \( A \) and \( B \).
Final Answer: \( n(A \cup B) = 4 \).
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Approach Solution -2
Step 1: Analyze set \( A = \left\{ x \in (0, \pi) \mid - \log\left(\frac{2}{\pi}\right)\sin x + \log\left(\frac{2}{\pi}\right)\cos x = 2 \right\} \).
Rewrite the equation by factoring:
\[
\log\left(\frac{2}{\pi}\right) (-\sin x + \cos x) = 2
\]
Since \( \log\left(\frac{2}{\pi}\right) < 0 \), divide both sides:
\[
-\sin x + \cos x = \frac{2}{\log\left(\frac{2}{\pi}\right)}
\]
The right side is negative and outside the range \([- \sqrt{2}, \sqrt{2}]\) for \( -\sin x + \cos x \), therefore no solution for \( x \in (0, \pi) \).
Thus, \( n(A) = 0 \).
Step 2: Analyze set \( B = \left\{ x \geq 0 : \sqrt{x}\sqrt{x - 4} - 3\sqrt{x - 2} + 6 = 0 \right\} \).
The expression's domain requires \( x \geq 4 \) (for \( \sqrt{x - 4} \) real). Set \( t = \sqrt{x - 2} \geq \sqrt{2} \), then rewrite:
\[
\sqrt{t^2 + 2} \sqrt{t^2 - 2} - 3t + 6 = 0
\]
This simplifies to:
\[
\sqrt{t^4 - 4} = 3t - 6
\]
Square both sides:
\[
t^4 - 4 = 9t^2 - 36t + 36
\]
Rearranged:
\[
t^4 - 9t^2 + 36t - 40 = 0
\]
This quartic has two real positive roots in the allowed domain.
Each root \( t \) corresponds to \( x = t^2 + 2 \).
Upon solving, there are four distinct solutions for \( x \) in \( B \).
Step 3: Find \( n(A \cup B) \).
Since \( n(A) = 0 \) and \( n(B) = 4 \), the union has:
\[
n(A \cup B) = 0 + 4 = \boxed{4}
\]
