Animated Solution for Mathematics - Trigonometry: Let $A=\{x\in (0,\pi )-\{\frac{\pi }{2}\}:{\log }_{(2/\pi )}|\sin x|+{\log }_{(2/\pi )}|\cos x|=2\}$ and $B=\{x\ge 0:\sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\}$. Then $n(A\cup B)$ is equal to :

We begin with the equation:

Using the logarithmic property ${\log }_{b}m+{\log }_{b}n={\log }_b(mn)$, we combine the terms:

Converting this to exponential form, we raise the base $(2/\pi )$ to the power of $2$:

To simplify, we multiply both sides by $2$ to utilize the double angle identity $2\sin x\cos x=\sin 2x$:

The problem specifies the domain $x\in (0,\pi )-\{\frac{\pi }{2}\}$. Multiplying this range by $2$, the domain for $2x$ becomes $(0,2\pi )-\{\pi \}$.

Since $\frac{8}{{\pi }^2}\approx 0.81$, a horizontal line at $y=0.81$ intersects the graph of $|\sin 2x|$ at exactly $4$ points within the interval $(0,2\pi )$. Thus, $n(A)=4$.

We consider the equation $\sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0$. Let $t=\sqrt{x}$, where $t\ge 0$. The equation becomes:

We analyze this using two cases based on the absolute value:

Case 1: $t\ge 2$ The equation becomes $t^2-4t-3(t-2)+6=0$, which simplifies to: Both $t=3$ and $t=4$ satisfy the condition $t\ge 2$.

Case 2: $0\le t<2$ The equation becomes $t^2-4t-3(2-t)+6=0$, which simplifies to: Both $t=0$ and $t=1$ satisfy the condition $0\le t<2$.

Squaring these values ($t^2=x$), we find $x\in \{0,1,9,16\}$. Thus, $n(B)=4$.

We have determined that $n(A)=4$ and $n(B)=4$. To find $n(A\cup B)$, we check for any intersection between the sets.

The values $x\in \{0,9,16\}$ are outside the domain of Set A, and $x=1$ does not satisfy the original logarithmic equation. Because the sets are disjoint, we simply add the number of elements:

The final result is $8$.