M={0丨sin0≥,0≤α≤π},N={0丨cosα≤,0≤α≤π},则M∩N=?3Q

问题描述:

M={0丨sin0≥,0≤α≤π},N={0丨cosα≤,0≤α≤π},则M∩N=?3Q

sinα cosα=2/3 (sinα)^2 (cosα)^2=1 所以2sinαcosα=-5/9 (sinα-cosα)^2=1-(-5/9)=14/9又α∈(0,π),所以sinα-cosα=√14/3 sinα cosα=2/3 由以上两个方程解得 sinα=(2 √14)/6 cosα=(2-√14)/6...