The function f (x) = 2Sin (2x + (π / 6)) + A is known, where a is a constant Finding monotone decreasing interval of F (x) When x ∈ [0, π / 2], the maximum value of F (x) is 4

The function f (x) = 2Sin (2x + (π / 6)) + A is known, where a is a constant Finding monotone decreasing interval of F (x) When x ∈ [0, π / 2], the maximum value of F (x) is 4

The function f (x) = 2Sin (2x + π / 6) + m + 1. Find the monotone increasing interval of F (x) on [0, π]

f(x)=2sin(2x+π/6)+m+1
Let t = 2x + π / 6, t ∈ [π / 6,2 π + π / 6]
f(t)=2sin(t)+m+1
The monotone increasing interval of F (T) is:
T ∈ [π / 6, π / 2] or t ∈ [3 π / 2,2 π + π / 6]
At this point:
X ∈ [0, π / 6] or X ∈ [2 π / 3, π]
That's what we want

The monotone increasing interval of function f (x) = 2Sin (x - π 3), X ∈ [- π, 0] is ()
A. [-π，-5π6]B. [-5π6，-π6]C. [-π3，0]D. [-π6，0]

Let 2K π - π 2 ≤ X - π 3 ≤ 2K π + π 2, K ∈ Z, the solution is 2K π - π 6 ≤ x ≤ 2K π + 5 π 6, ∵ x ∈ [- π, 0], f (x) = 2Sin (x - π 3), and the monotone increasing interval of X ∈ [- π, 0] is [- π 6, 0]