Find the maximum value of the function y = (SiNx) ^ 2 + acosx + 5 / 8a-3 / 2 (0 ≤ x ≤ π / 2)

Find the maximum value of the function y = (SiNx) ^ 2 + acosx + 5 / 8a-3 / 2 (0 ≤ x ≤ π / 2)

There are three cases: when 1,0 ≤ a ≤ 2, the maximum value is a ^ 2 / 4 + 5A / 8-1 / 2
2, A2, the maximum is 13A / 8-1 / 2

Given that the definition field of function f (x) = 12cos2x + asinx − A4 is [0, π 2], and the maximum value is 2, find the value of real number a

The definition domain of the function is [0, π [0 [0, π 2] for [0, π 2] and [0, π 2] for [12 (1 − 2s2x2x) + 12 (1 − 2s2x2x2x) + asinx {12 (x (x) = 12cos2x (12cos2x) + 12cos2x + asinx cos2x + asinx {12 (12) 2x + asinx2x x x} = 12 (12, the definition of [12, π [0, π 2 [12, π [12, π [12 {A4 = 12} A4 = 12} 4 = 12 \\\\\\\\\\\\\\thevalue is - 1 + A + 12 − A4 = 2 & nbsp; & nbsp; In conclusion, a = - 6 or a = 53

It is known that the function f (x) = the square of 2A (cosx) + bsinxcosx, and f (0) = 2, f (π / 3) = 1 / 2 + 3 / 2
(1) Find the minimum positive period of function f (x)
(2) Find the maximum value of function f (x) and the set of X when the maximum value is obtained
(3) Find the decreasing interval of function f (x) on [0, π]

If f (x) is reduced first, then
f(x)
=a*(1+cos2x)+b*(1/2)*sin2x
=a+a*cos2x+(b/2)*sin2x
∵f(0)=2
A + a = 2, a = 1
∵f(π/3)=1/2+√3/2
∴a-(a/2)+b*√3/4=1/2+b*√3/4
=1/2+√3/2
∴b=2
∴f(x)
=1+cos2x+sin2x
=1+√2*sin(2x+π/4)
(1) The minimum positive period is
T=2π/2=π
(2) It's easy to know
What is the maximum value of F (x)
1+√2
here
2x+π/4=π/2+2kπ,k∈Z
The set of values of X is
{x|x=π/8+kπ,k∈Z}
(3) The monotone decreasing interval of F (x) is
π/2+2kπ