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?

You need to solve $\cos \left(2 \arcsin(-x) \right) = 0$. Let $y = 2 \arcsin(-x)$ then $\cos y  = 0$ so $y = \pi/2 \pm n\pi$. Then,
$$
2 \arcsin(-x) = \frac{pi}{2} \pm n\pi
$$
which implies
$$
x = -\sin \left( \frac{\pi}{4} \pm \frac{n\pi}{2} \right)
$$
Can you simplify this?
