dy/dx when x = cos(y)

Can someone give me a hint on what to do next in the following question:

If $x = cos y$ find $\frac{dy}{dx}$. Give your answer in terms of x.

$\frac{dx}{dy} = -sin y$
$\frac{dy}{dx} = -csc y = -csc(arccos x)$

This is where I'm stuck.

[Edit: Found a solution elsewhere]

You are right
$\frac{dy}{dx} = -csc y = -csc(arccos x)=-\frac{1}{sin(arccosx)}=-\frac{1}{\sqrt{1-(cos(arccosx))^2}}=-\frac{1}{\sqrt{1-x^2}}$

So you differenciated y=arccosx from its x=cosy implicit form
which is known as differentiation of inverse function.
$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$
You can get the derivative of f.e arcsinx, arctanx, lnx, arsinhx, artanhx with this method (going back to a known derivative with invertation) and memorize them as base differentiation rules.