The number of solutions of $sin \begin{Bmatrix} sin^{-1}\left(log_{1/2}x\right)\end{Bmatrix} + 2\begin{vmatrix}cos \begin{Bmatrix}sin^{-1}\left(\frac{x}{2}-\frac{3}{2}\right)\end{Bmatrix}\end{vmatrix}=0$, is

2

$\sin(\sin^{-1}(\log_{1/2}x))=\log_{1/2}x=-\log_2 x$

$\cos\left(\sin^{-1}\left(\frac{x-3}{2}\right)\right)=\sqrt{1-\left(\frac{x-3}{2}\right)^2}$

$\Rightarrow -\log_2 x + 2\sqrt{1-\left(\frac{x-3}{2}\right)^2}=0$

$2\sqrt{1-\left(\frac{x-3}{2}\right)^2}=\log_2 x$

$\text{Domain: } -1 \le \frac{x-3}{2} \le 1 \Rightarrow 1 \le x \le 5,\ \text{and } x>0$

$\text{Also } \log_2 x \ge 0 \Rightarrow x \ge 1$

$\text{Square both sides:}$

$4\left(1-\frac{(x-3)^2}{4}\right)=(\log_2 x)^2$

$4-(x-3)^2=(\log_2 x)^2$

$\text{Check endpoints:}$

$x=1:\ 4-( -2)^2=0,\ (\log_2 1)^2=0 \Rightarrow \text{solution}$

$x=5:\ 4-(2)^2=0,\ (\log_2 5)^2\ne0 \Rightarrow \text{reject}$

$\text{LHS }=4-(x-3)^2 \text{ is symmetric about } x=3,\ \text{RHS increases for } x>1$

$\text{At } x=3:\ \text{LHS}=4,\ \text{RHS}=(\log_2 3)^2<4$

$\text{At } x=5:\ \text{LHS}=0,\ \text{RHS}>0$

$\text{Thus one more intersection in } (3,5)$

$\text{Total solutions }=2$

$\text{Number of solutions} = 2$