Let t=cosx∈[- 1, 1]

y=2t？ -2at-(2a+ 1)

=2(t？ -At +a? /4)-(a？ /2+2a+ 1)

=2(t-a/2)？ -(a? /2+2a+ 1)

This is a quadratic function about t, but t∈[- 1, 1]

To find the expression of the minimum value of a function, it is necessary to discuss the symmetry axis.

Three positional relationships between t=a/2 and interval [- 1, 1].

1) When A/2 ≤- 1, that is, A ≤- 2, [- 1, 1] is on the right side of the symmetry axis, the function increases.

∴t=- 1, and y takes the minimum value f(a)=y|(t=- 1)= 1.

2) When-1

The function decreases on [- 1, a/2] and increases on [a/2, 1].

∴t=a/2, y takes the minimum value f(a)=y|(t=a/2)=-a? /2-2a- 1

3) When a/2≥ 1, that is, a≥2, [- 1, 1] is on the left side of the symmetry axis, the function decreases.

∴t= 1, and y takes the minimum value f (a) = y | (t =1) =1-4a.

To sum up, the function minimum expression f(a) is segmented:

{ 1，(a≤-2)

f(a)={-a？ /2-2a- 1，(-2 & lt； a & lt2)

{ 1-4a，(a≥2)

(2)

If f(a)= 1/2, only -2.

∴-a？ /2-2a- 1= 1/2

Answer? The solution of +4a+3=0 gives a=- 1 or a=-3 (truncation).

∴a=- 1

Y=2t at this time? +2t+ 1，

When t= 1, y takes the maximum value of 5.