<s>[INST] 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?
 [/INST] Let $\sin^{-1}(-x)=u\implies\sin u=-x$
$$\cos(2\sin^{-1}(-x))=\cos2u=1-2\sin^2u=1-2(-x)^2$$
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