Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$ 
Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$

I get the answer $\frac{-1}{ \sqrt2}$
By solving like this 
\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\
2\sin^{-1}(- x) &= \frac\pi2\\
\sin^{-1}(- x) &= \frac\pi4\\
-x &= \sin\left(\frac\pi4\right)\end{align}
Thus $x =\frac{-1}{\sqrt2}$
But correct answer is $\pm\frac{1}{\sqrt2}$
Where am I going wrong?

Okay so expand the double angle to get $\cos^2(\sin^{-1}(-x))-\sin^2(\sin^{-1}(-x)=0$.
This should give you $(1-(-x)^2)-(-x)^2=0$.
Finally you have $1-2x^2=0$.
Solutions are $\displaystyle \boxed{\pm \frac{1}{\sqrt{2}}}$.
