$\text { Show that } \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=-2 \pi+2 \cos ^{-1} x \text { if }-1 \leq x \leq-\frac{1}{\sqrt{2}} $ 
Show that
$$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=-2 \pi+2 \cos ^{-1} x$$ if $-1 \leq x \leq-\frac{1}{\sqrt{2}}.$

I tried solving this question a lot but I’m unable to. My answer comes different.
$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ Putting $x=\cos \theta$
$$
\begin{array}{l}
=\sin ^{-1}\left(2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right) \\
=\sin ^{-1}\left(2 \cos \theta \sqrt{\sin ^{2} \theta}\right) \quad\left(\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right)
\end{array}
$$
$$
=\sin ^{-1}(2 \cos \theta \sin \theta)
$$
$$
=\sin ^{-1}(2 \cos \theta \sin \theta)
$$
$$
=-\sin ^{-1}(\sin 2\theta)
$$
(Using $\sin 2 x=2 \sin x \cos x$)
$=2 \theta \quad$ As $x=\cos \theta$
$=2 \times \cos ^{-1} x$
$=2 \cos ^{-1} x$
EDIT: If any confusion with Q writing


hint
Let
$$f(x)=\arcsin(2x\sqrt{1-x^2})-2\arccos(x)$$
assuming $ f $ is differentiable at $ [-1,-\frac{1}{\sqrt{2}}] $,
prove that
$$f'(x)=0$$
So
$$f(x)=Cte=f(-1)=-2\pi$$
Other approach
Put $$x=\cos(\frac{\theta}{2})$$
