Let A=x∈(0,π)−π2: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 A=x∈(0,π)−π2: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

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Let $A = \{x \in (0, π) - \{\frac{π}{2}\} : \log_{(2 / π)} | \sin x | + \log_{(2 / π)} | \cos x | = 2\}$ 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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Detailed Solution

$\begin{array}{l} A : \log_{2 π} | \sin x | + \log_{2 π} | \cos x | = 2 \\ \Rightarrow \log_{2 π} (| \sin x \cdot \cos x |) = 2 \\ \Rightarrow | \sin 2 x | = \frac{8}{π^{2}} \end{array}$
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B : Let $\sqrt{x} = t < 2$
Then $\sqrt{x} (\sqrt{x} - 4) + 3 (\sqrt{x} - 2) + 6 = 0$
$\begin{array}{l} \Rightarrow t^{2} - 4 t + 3 t - 6 + 6 = 0 \\ \Rightarrow t^{2} - t = 0, t = 0, t = 1 \\ x = 0, x = 1 \end{array}$
again let $\sqrt{x} = t > 2$
then $t^{2} - 4 t - 3 t + 6 + 6 = 0$
$\begin{array}{l} \Rightarrow t^{2} - 7 t + 12 = 0 \\ \Rightarrow t = 3, 4 \\ x = 9, 16 \end{array}$
Total number of solutions
$n (A \cup B) = 4 + 4 = 8$