19: Denoting h(x) = sin(x) − 2x/π one gets

h′(x) = cos(x) − 2/π.

This is equal to zero only once within the interval M = (0,π/2), at some c, so we know that h′ is positive on (0,c) and negative on (c,π/2).

Consequently h is increasing on [0,c] and decreasing on [c,π/2]. Putting this together with the fact that h(0⁺) = 0 and h((π/2)^-) = 0 we obtain that h > 0 on (0,π/2) as required.