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

Given, (sin⁡x)x3−2x+b≥2Let f(x)=(sin⁡α)x2−2x+b−2Abscissa of the vertex is given by x=1sin ⁡α>1The graph of f(x)=(sin⁡α)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,sin⁡α−2+b−2≥0,b≥4−sin⁡α,α∈(0,π)