If 0≤x≤2π and 2cosec⁡2x12y2−y+1≤2 then number of ordered pairs of (x,y) is

2cosec2x12y2−y+1≤2⇒2cosec2xy2−2y+2≤22cosec2x(y−1)2+1≤2----iWe have∴2cosec 2x≥2 and ⇒(y−1)2+1≥1⇒ product ≥2----iiFrom (i) and (ii)2cosec⁡x2x=2 and (y−1)2+1=1⇒ sin⁡x=±1 and y=1⇒ x=π2,3π2 and y=1Hence, the solution of the given inequality is (x,y)=π2,1,3π2,1