Solve $\cot^2x=\csc x$ in degrees $$\cot^2x = \csc x$$
I have to solve for $x$ in degrees. Here's what I did:
$$\cot^2x=\csc^2x - 1$$
$$\csc^2x -\csc x - 1 = 0$$
If $y=\csc x$:
$$y^2-y-1=0$$
and now I cannot proceed to the given answers of 38.2° and 141.2°.

since $$\frac{\cos(x)^2}{\sin(x)^2}=\frac{1}{\sin(x)}$$ you will get $$\cos(x)^2=\sin(x)$$ this means $$1-\sin(x)^2-\sin(x)=0$$
and so you will get a quadratic equation in $$\sin(x)$$ 
