Find the function y = sin ^ 2x + acosx-5a / 8-3 / 2 (0

Find the function y = sin ^ 2x + acosx-5a / 8-3 / 2 (0

y= sin^2x+acosx-5a/8-3/2=-cos^2x+acosx-5a/8-1/2
Let cosx = t, the value range of t be [0,1]
Then y = - T ^ 2 + at-5a / 8-1 / 2, its symmetry axis is t = A / 2
Discussion:
(1) If a / 2 ≥ 1, the maximum value of Y is the value of T = 1, the solution is a = 20 / 3
(2) If 0

Given the function f (x) = sin ^ 2x + acosx + 5A / 8-3 / 2, a ∈ R. when a = 1, find the maximum value of function f (x)

F (x) = 1-cos & sup2; X + cosx + 5 / 8-3 / 2, let t = cosx, t ∈ [- 1,1]
In this way, y = 1-T & sup2; + T + 5 / 8-3 / 2, which is an interval maximum problem of quadratic function

Is there a real number a such that the maximum value of the function f (x) = sin ^ 2x + acosx + 5 / 8A in the interval [0, π / 2] is 5 / 2? If so, the value of a is obtained

1.5
F (x) = - (cosx - (A / 2)) + square A / 4 + 5A / 8 + 1
(1) When a / 2 is less than 0, the maximum value of F (x) = f (x = π / 2) = 5A / 8 + 1 = 5 / 2, a = 12 / 5 is obtained, which is contradictory to a / 2 less than 0, and is rounded off
(2) When a / 2 is greater than 1, the maximum value of F (x) = f (x = 0) = 13A / 8 = 5 / 2, a = 20 / 13 is contradictory to a greater than 2
(3) When 0 ″ A / 2 ″ 1, the maximum value of F (x) = f (cosx = A / 2) = [2 * (a) + 5A + 8] / 8 = 5 / 2, a = 1.5 or - 4 (contradicting with 0 ″ a ″ 2, rounding off), a = 1.5 is in line with the meaning of the question
In conclusion, a = 1.5
The results show that: (1) when cosx is a decreasing function in a quadrant, the opening of quadratic function is downward, and the variable t (taking cosx as a whole T) is on the right side of the symmetry axis, it is a decreasing function. In general, the compound function f (x) is an increasing function, X is the largest, f (x) is the largest; similarly, we speculate on (2) (3) conditions

If the maximum value of function y = sin2x + acosx-1 / 2a-3 / 2 is 1, find the value of A. the answer is 1-radical 7 or 5
Could you please write down the results and the previous ones in detail

sin2x+acosx-1/2a-3/2
Derivation
2cos2x-asinx
Let it be zero, then find the maximum value at this point
It can be changed into:
4(cosx)^2-asinx-2=0
Then 4-4 (SiNx) ^ 2-asinx-2 = 0
The solution is SiNx = a + radical (a ^ 2-32) / 8, so cosx = radical {1 - [a + radical (a ^ 2-32) / 8] ^ 2}
Set SiNx = a + radical (a ^ 2-32) / 8
It can be obtained by substituting sin2x + acosx-1 / 2a-3 / 2