If u derivate the function f(x) u will have a function [f'(x)] that will give u the slope for any value of x:

so, the derivate of the function \(f(x)=2cos(x) is f'(x) = -2sin(x)\), now to evalute where the derivate is equal to 1 (the slope that are asking for).

\(1 = -2 \sin(x) -> \sin^{-1} \frac{-1}{2} = \frac{-\pi}{6} = -0.53\)

So the x,y coordinate are (-0.53 , 1.72)

so, the derivate of the function \(f(x)=2cos(x) is f'(x) = -2sin(x)\), now to evalute where the derivate is equal to 1 (the slope that are asking for).

\(1 = -2 \sin(x) -> \sin^{-1} \frac{-1}{2} = \frac{-\pi}{6} = -0.53\)

So the x,y coordinate are (-0.53 , 1.72)