$\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

 A: 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 $$x=\cos(\frac{\theta}{2})$$
A: Continuing from the substitution in the question, where OP put $x=\cos \theta$. Here I pick $\theta$ to be the particular value $\frac{3\pi}4\le \theta\le \pi$, so that $\theta$ is the principal value $\theta = \cos^{-1}x$.
Using the formula to find general solutions of inverse $\sin$, if given $\sin 2\theta$, $2\theta$ would be in the form
$$2\theta = n\pi + (-1)^n\sin^{-1}(\sin 2\theta)$$
For this case where $\frac{3\pi}2\le 2\theta\le 2\pi$, so $n=2$,
$$\begin{align*}
2\theta &= 2\pi + \sin^{-1}(\sin 2\theta)\\
\sin^{-1}(\sin 2\theta)&=-2\pi + 2\theta\\
\sin^{-1}\left(2x\sqrt{1-x^2}\right) &= -2\pi + 2\cos^{-1}x
\end{align*}$$
The conversion from $\sin 2\theta = 2\cos\theta\sin\theta = 2x\sqrt{1-x^2}$ is what OP already knows.
($0 \le \sin \theta \le 1$ is important here, because that confirms that $\sqrt{\sin^2\theta} = \left|\sin\theta\right| = \sin\theta$.)
