The number of distinct solutions of the equation,
${\log }_{\frac{1}{2}}|\sin x|=2-{\log }_{\frac{1}{2}}|\cos x|$ $\text{in the interval}[0,2\pi ],\text{is}$

$\text{Given}:{\log }_{\frac{1}{2}}|\sin x|=2-{\log }_{\frac{1}{2}}|\cos x|,x\in [0,2\pi ]$

$\Rightarrow {\log }_{\frac{1}{2}}|\sin x|+{\log }_{\frac{1}{2}}|\cos x|=2$

$\Rightarrow {\log }_{\frac{1}{2}}|\sin x||\cos x|=2$

$\Rightarrow |\sin x\cos x|={\left(\frac{1}{2}\right)}^2$

$\Rightarrow \frac{|2\sin x\cos x|}{2}=\frac{1}{4}$

$\Rightarrow |\sin 2x|=\frac{1}{2}$

$\therefore \sin 2x=\pm \frac{1}{2}$


$\therefore$ We have 8 solutions for $x\in [0,2\pi ]$