$$y = sin x, -\pi \leq 0 \leq \pi, y' = cos x, y'' = -sin x$$

$$\kappa = \frac{|-sin x|}{(1 + cos^2 x)^{\frac{3}{2}}}$$

Considering x only from 0 to pi lets me remove the absolute value bars.

$$\frac{d\kappa}{dx} = \frac{cos x}{(1 + cos^2 x)^{\frac{3}{2}}} + \frac{3cos x sin^2 x}{(1 + cos^2 x)^{\frac{5}{2}}}$$

$$\frac{d\kappa}{dx} = 0 \longrightarrow \frac{cos x}{(1 + cos^2 x)^{\frac{3}{2}}} = -\frac{3cos x sin^2 x}{(1 + cos^2 x)^{\frac{5}{2}}} \longrightarrow 1 + cos^2 x = -3sin^2 x = -2sin^2 x -sin^2 x \longrightarrow 2 = -2 sin^2 x \longrightarrow sin^2 x = -1$$

This gives me an imaginary root. Wolfram tells me this function has real roots: https://www.wolframalpha.com/input/?...sx)%5E2)%5E1.5)