The function y = 2Sin (2x − π) 6) The monotone decreasing interval of is______ .

Let 2x − π
6∈[π
2+2kπ，3π
2 + 2K π], X ∈ [K π + π] can be obtained
3，  kπ+5π
6](k∈Z)
The function y = 2Sin (2x − π)
6) The monotone decreasing interval is [K π + π]
3，  kπ+5π
6](k∈Z)
So the answer is: [K π + π
3，  kπ+5π
6](k∈Z)

Function y = Log1 The interval of 2 x 2 is monotone______ .

According to the meaning of the title, the definition domain of the function is: (2, + ∞) ∪ (∞, 0)
Let t = x2-2x, then y = Log1
2T
Because the function y = Log1
2T decreases monotonically on the domain of definition
T = x2-2x increases monotonically at (2, + ∞) and decreases monotonically at (- ∞, 0)
According to the monotonicity of the composite function, the function y = Log1
The monotone decreasing interval of 2 (x2 − 2x) is: (2, + ∞)
So the answer is: (2, + ∞)

Function y = Log1 The interval of 2 x 2 is monotone______ .

The function y = 2Sin (2x + π) 6) The monotone decreasing interval of (x ∈ [- π, 0]) is___ .

∵ the monotone decreasing interval of sine function is [- 3 π]
2，-π
2]，
∴-3π
2≤2x+π
6≤-π
And [- x, π],
The solution is - 5 π
6≤x≤-π
3，
Then the monotone decreasing interval of the function is [- 5 π
6，-π
3]．
So the answer is: [- 5 π
6，-π
3]．

Fill in the blanks! The function y = 2Sin (2x + Pai / 6) x belongs to the monotone decreasing interval of [- Pai, 0] Ask minus interval

∵y=2sin(2x+π/6)
∴y'=4cos(2x+π/6)
∵ x ∈ [- π, 0], that is - π ≤ x ≤ 0
∴-11π/6≤2x+π/6≤π/6
∵ when - π

Function 2Sin (π / 6-2x), X ∈ (0, π) is the interval of increasing function

y=2sin(π/6-2x)=-2sin(2x-π/6)
Because of the minus sign in front of it, π / 2 + 2K π

The function y = 2Sin (2x − π) 6) The monotonically increasing interval of is______ .

Let 2K π - π
2≤2x-π
6≤2kπ+π
2, K ∈ Z, K π - π
6≤x≤kπ+π
3，
So the function y = 2Sin (2x − π)
6) The monotone increasing interval of is [K π - π
6，kπ+π
3]，k∈z，
So the answer is [K π - π
6，kπ+π
3]，k∈z．

Given the function f (x) = cos (π / 2 + 2x) + cos (2x) 1), find the monotone decreasing interval of function y = f (x) 2) If f (α - π / 8) = radical 2 / 3, find the value of F (2 α + π / 8)

1、f(x)=-cos(π-π/2-2x)+cos2x
=-cos(π/2-2x)+cos2x
=cos2x-sin2x
=√2[cos2x(√2/2）-sin2x（√2/2）]
=√2cos(2x+π/4）,
2K π < = 2x + π / 4 < = 2K π + π,
Then 2K π - π / 4 < = 2x < = 2K π + 3 π / 4,
ν K π - π / 8 < = x < = k π + 3 π / 8, (K ∈ z), is a monotonic decreasing interval
2、f（α-π/8）=√2cos(2α-π/4+π/4）=√2cos2α=√2/3,
cos2α=1/3,
sin 2α=±√（1-1/9）=±2√2/3,
2 α∈ [2K π, 2K π + π] is positive,
That is, α∈ [K π, K π + π / 2] is positive,
α∈ [K π - π / 2, K π] is negative,
f（2α+π/8）=√2cos[4α+π/4+π/4）
=√2cos[4α+π/2）
=-√2cos（π-4α-π/2）
=-√2cos(π/2-4α）
=-√2sin4α
=-√2*2sin（2α）cos（2α）
=-2√2（±2√2/3）*（1/3）
=±8/9.
α∈ [K π, K π + π / 2] is negative,
α∈ [K π - π / 2, K π] is positive

The function y = 2Sin (2x − π) 6) The monotone decreasing interval of is______ .