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

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Question

$	ext { A: } \log _{(2 / \pi)}|\sin x|+\log _{(2 / \pi)}|\cos x|=2$
$\Rightarrow \log _{(2 / \pi)}(|\sin x \cdot \cos x|)=2$
$\Rightarrow|\si

Vedclass · Accepted Answer

$8$

Vedclass · Answer

$	ext { A: } \log _{(2 / \pi)}|\sin x|+\log _{(2 / \pi)}|\cos x|=2$
$\Rightarrow \log _{(2 / \pi)}(|\sin x \cdot \cos x|)=2$
$\Rightarrow|\sin 2 x|=\frac{8}{\pi^2}$
image
Number of solution 4
$B$ : let $\sqrt{ x }= t <2$
Then $\sqrt{x}(\sqrt{x}-4)+3(\sqrt{x}-2)+6=0$
$\Rightarrow t^2-4 t+3 t-6+6=0$
$\Rightarrow t^2-t=0, t=0, t=1$
$x=0, x=1$
again let $\sqrt{ x }= t >2$
then $t^2-4 t-3 t+6+6=0$
$\Rightarrow t^2-7 t+12=0$
$\Rightarrow t=3,4$
$x=9,16$
Total number of solutions
$n(A \cup B)=4+4=8$

Let $A=\left\{x \in(0, \pi)-\left\{\frac{\pi}{2}\right\}: \log _{(2 / \pi)}|\sin x|+\log _{(2 / \pi)}|\cos x|=2\right\}$ 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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