Let A = {x in (0, pi) - {frac{pi}{2}} : log_{ | vedclass

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(C) For set $A$: $\log_{(2/\pi)}|\sin x| + \log_{(2/\pi)}|\cos x| = 2$$\Rightarrow \log_{(2/\pi)}|\sin x \cos x| = 2$$\Rightarrow |\sin x \cos x| = (2

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$8$

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(C) For set $A$: $\log_{(2/\pi)}|\sin x| + \log_{(2/\pi)}|\cos x| = 2$$\Rightarrow \log_{(2/\pi)}|\sin x \cos x| = 2$$\Rightarrow |\sin x \cos x| = (2/\pi)^2 = 4/\pi^2$$\Rightarrow |\frac{1}{2} \sin 2x| = 4/\pi^2$$\Rightarrow |\sin 2x| = 8/\pi^2$Since $8/\pi^2 \approx 8/9.86 For set $B$: Let $t = \sqrt{x} \geq 0$. The equation is $t(t - 4) - 3|t - 2| + 6 = 0$.Case $1$: $0 \leq t $t^2 - 4t - 3(2 - t) + 6 = 0$ $\Rightarrow t^2 - 4t - 6 + 3t + 6 = 0$ $\Rightarrow t^2 - t = 0$ $\Rightarrow t(t - 1) = 0$.Since $0 \leq t Case $2$: $t \geq 2$. Then $|t - 2| = t - 2$.$t^2 - 4t - 3(t - 2) + 6 = 0$ $\Rightarrow t^2 - 4t - 3t + 6 + 6 = 0$ $\Rightarrow t^2 - 7t + 12 = 0$ $\Rightarrow (t - 3)(t - 4) = 0$.Thus $t = 3$ and $t = 4$. So $x = 3^2 = 9$ and $x = 4^2 = 16$.Set $B = \{0, 1, 9, 16\}$,so $n(B) = 4$.Since sets $A$ and $B$ are disjoint,$n(A \cup B) = n(A) + n(B) = 4 + 4 = 8$.

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

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