F (x) = SiNx ^ 2 + acosx + 5 / 8a-3 / 2, a belongs to R (1). When a = 1, find the maximum value of function f (x) (2) if for the function on interval [0, π / 2] (1) When a = 1, find the maximum value of function f (x) (2) If f (x) ≤ 1 holds for any X in the interval [0, π / 2], the value range of a is obtained

F (x) = SiNx ^ 2 + acosx + 5 / 8a-3 / 2, a belongs to R (1). When a = 1, find the maximum value of function f (x) (2) if for the function on interval [0, π / 2] (1) When a = 1, find the maximum value of function f (x) (2) If f (x) ≤ 1 holds for any X in the interval [0, π / 2], the value range of a is obtained

(1) Let t = cosx, the range of T is [- 1,1], f (x) max = 3 / 8;
(2) The range of a is a > = 1 / 2

If the image of the function y = sin2x + acosx is symmetric with respect to the line x = - π / 8, then a=

y=sin2x+acosx
=2cos^2-1+acosx
=2cosx^2+acosx-1
Axis of symmetry x = - A / 4
And because the image is symmetric with respect to a straight line, x = - π / 8
So, a = π / 2

If the image of F (x) = sin2x + acosx is symmetric with respect to the line x = - Π / 8, then a=_____
If it's like this on the first floor, I'll do it

This is a question to fill in the blanks, which can be done according to the definition of symmetry. The image is symmetrical about the straight line x = - Π / 8, that is, f (- Π / 8 + x) = f (- Π / 8-x) satisfies all x values
Let's go to a special value (easy to calculate) and bring it in to get the value of A
Here I take x = Π / 8, then:
f(-∏/8+x)=f（0）=a
F (- Π / 8-x) = f (- Π / 4) = - 1 + under a * radical (1 / 2)
F (- Π / 8 + x) = f (- Π / 8-x) get a = - (2 + radical 2)