The number of solutions of the equation $\sin \left[2 \cos^{-1} \left\{\cot \left(2 \tan^{-1} x\right)\right\}\right] = 0$ is

Let $f(x) = \cos \left(2 \tan ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1-x}{x}}\right)\right)$ for $0 < x < 1$. Then :

Let $E_1 = \{x \in R : x \neq 1 \text{ and } \frac{x}{x-1} > 0\}$ and $E_2 = \{x \in E_1 : \sin^{-1}(\log_e(\frac{x}{x-1})) \text{ is a real number}\}$. (Here,the inverse trigonometric function $\sin^{-1} x$ assumes values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$). Let $f : E_1 \rightarrow R$ be the function defined by $f(x) = \log_e(\frac{x}{x-1})$ and $g : E_2 \rightarrow R$ be the function defined by $g(x) = \sin^{-1}(\log_e(\frac{x}{x-1}))$. Match the items in $LIST I$ with $LIST II$.

$LIST I$	$LIST II$
$P$. The range of $f$ is	$1$. $(-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty)$
$Q$. The range of $g$ contains	$2$. $(0, 1)$
$R$. The domain of $f$ contains	$3$. $[-\frac{1}{2}, \frac{1}{2}]$
$S$. The domain of $g$ is	$4$. $(-\infty, 0) \cup (0, \infty)$
	$5$. $(-\infty, \frac{e}{e-1}]$
	$6$. $(-\infty, 0) \cup (\frac{1}{2}, \frac{e}{e-1}]$

If $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$, then $x \in$

The total number of real solutions of the equation $\theta=\tan ^{-1}(2 \tan \theta)-\frac{1}{2} \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^2 \theta}\right)$ is $($Here,the inverse trigonometric functions $\sin ^{-1} x$ and $\tan ^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,respectively.$)$

The number of solutions of $\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x$ is