f'(x)= 1-asinx

aE(0. 1) SinxE[-l，l]asinX & lt； 1

f’(x)>0

F(x) is increasing.

Let x=b-2 and f (b-2) =-2+acos (b-2) < 0.

f(b+2)>0

f(b-2)f(b+2)& lt； 0

According to the zero theorem, there must be a solution when xE[b-2, b+2].

And f(x) is increasing, so there is a unique solution.

Attached; y=f(x)

Y'>0, increased; Y'<0, negative

Y''>0 concave. Conversely, convex.

For f' () = 0

When f "() > 0, it is concave at the extreme point and has a minimum value f ().

On the contrary, on the contrary.

When f'' () = 0, if f'' () is not 0, it is an inflection point.