Let A ={x ∈(0, π)-{(π)/2}: log_(2 / π)| sin x|+ log_(2 / π)| cos x|=2} and B ={x ≥ 0: √ x (√ x -4)-3|√ x -2|+6=0}. Then n ( A ∪ B ) is equal to:

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