If (sinα)x2−2x+b≥2,  for all real values of x≤1  and α∈(0, π/2)∪(π/2, π),  then possible real values of b  is/are

Given, (sin α)x2−2x+b≥2. Let f(x)=(sinα)x2−2x+b−2 Abscissa corresponding to the vertex is given by             x=1sinα>1             The graph of f(x)=(sinα)x2−2x+b−2, ∀  x≤1,   is shown in the figure. Therefore, minimum of f(x)=(sin α)x2−2x+b−2  must be greater than zero but minimum is at x=1.  That is,  sinα−2+b−2≥0, b≥4−sinα, α∈(0,π)