Find an equation of the tangent line to the graph of the function at the given point

Find an equation of the tangent line to the graph of the function at the given point.
y = 1/3 arccos x

I know the answer is -(sqrt2/3)x+pi/12+1/3 I want to know how to get it...

I got y'=(1/3)(-1/sqrt1-x^2)
Then what do I do?

The point is not given. If it were, you would just sub the x-value of the point into the derivative and solve.