Question

Let f(x)=cot(x). Determine the points on the graph of f for 0<x<2π where the tangent line(s) is (are) parallel to the line y=−2x.

so i've derived f(x) = cot(x) and have gotten f'(x)=-csc^2(x) which then I've made f'(x)=-2x
so now i have -2x=-csc^2(x)
or
2x=csc^2(x) but I am not sure how to proceed

Answer 1

but if y = -2 x
slope = dy/dx = -2

2 = csc ^2 x

Answer 2

ok so we've corrected that mistake, but what can I do after? do i then say sqrt(2)=csc(x) and then what?