The function f (x) = sin (x + α) + √ 3cos (x - α) is known, where 0 ≤ α < π and f (x) = f (- x) is constant for any real number X (1) Finding the value of α (2) Finding the maximum value and monotone increasing interval of function f (x)

The function f (x) = sin (x + α) + √ 3cos (x - α) is known, where 0 ≤ α < π and f (x) = f (- x) is constant for any real number X (1) Finding the value of α (2) Finding the maximum value and monotone increasing interval of function f (x)

(1)f(x)=sin(x+α)+√3cos(x-α)
=sinxcosa+cosxsina+√3（cosxcosa+sinxsina)
=sinx(cosa+√3sina)+cosx(sina+√3cosa),
From F (x) = f (- x), cosa + √ 3sina = 0, Tana = - 1 / √ 3,
0≤α＜π,
∴a=5π/6.
(2)f(x)=-cosx,
Its maximum value is 1, monotone increasing interval is [2K π, (2k + 1) π], K ∈ Z

Given that f (x) = 3sin (2x - π / 6), if there exists α∈ [0, π / 2], so that f (α + x) = f (2 α - x) is constant for all real numbers x, then α =. Process!

F (α + x) = f (2 α - x), so 3sin (2x + 2 α - π / 6) = 3sin (4 α - 2x - π / 6) = - 3sin [2x - (4 α - π / 6)]
SO 2 α - π / 6 + 4 α - π / 6 = π α = 2 π / 9

Given that a belongs to R, if the function f (x) = cosa * x2-4sina * x 6 belongs to R, take a positive value for all x, and find the value range of A

If Cosa ＞ 0, - 4sina + 6 ＞ 0, a is an internal angle of a triangle, and from Cosa ＞ 0, we know that a is an acute angle, and Sina ≤ 1, then the range of angle a is [0, π / 2]