Let y = f (x) be a decreasing function if θ∈ [0, π] 2] When f (COS 2 θ + 2msin θ) + F (- 2m-2) > 0 is constant, find the value range of M

Let y = f (x) be a decreasing function if θ∈ [0, π] 2] When f (COS 2 θ + 2msin θ) + F (- 2m-2) > 0 is constant, find the value range of M

From the condition, we can get: F (Cos2 θ + 2msin θ) > - f (- 2m-2) because y = f (x) is an odd function, there is f (- 2m-2) = - f (2m + 2) (2 points), that is, f (Cos2 θ + 2msin θ) > F (2m + 2) and because y = f (x) is a decreasing function, it is equivalent to Cos2 θ + 2msin θ < 2m + 2

Let f (x) be a decreasing function defined on (- ∞, 2], and f (a? - sinx-1) ≤ f (a + cos? X) holds for all x ∈ R

According to the theme
f（a²-sinx-1）≤f（a+cos²x）
Available
There is no solution for a 2 - sinx-1 < A + cos? X, that is, there is no such X
The result is a-a-1

The definition domain of the function f (x) = LG (sin2x cos2x) is______ .

From the meaning of the title: sin2x cos2x > 0,
That is, cos2x-sin2x < 0. From the formula of double angle, cos2x < 0 can be obtained,
So π
2+2kπ＜2x＜3π
2+2kπ，k∈Z，
∴kπ+π
4＜x＜kπ+3π
4，k∈Z，
So the answer is: {x| K π + π
4＜x＜kπ+3π
4，k∈Z}

Definition domain of function f (x) = ㏒ 10 (sin? X-cos? X)

sin²x-cos²x>0
cos²x-sin²x

Let f (x) = cos (2x + Pai / 3) + (sin ^ 2) x 1: find the range and minimum positive period of F (x). 2: let a, B, C Let me ask you a math problem: Let f (x) = cos (2x + Pai / 3) + (sin ^ 2) X 1: Find the range of F (x) and the minimum positive period 2: Let a, B and C be the three inner angles of △ ABC, and their opposite side lengths are ABC respectively. If cos C = (2 √ 2) / 3, a is an acute angle and f (A / 2) = (- 1 / 4) a + C = 2 + 3 √ 3, calculate the area of ABC Two questions,

F (x) = 1 / 2cos2x - √ 3 / 2sin2x + 1 / 2-1 / 2cos2x = - √ 3 / 2sin2x, so the value range is [- √ 3 / 2, √ 3 / 2] and the minimum positive period is π
A is an acute angle, f (A / 2) = - √ 3 / 2sina = 2 + 3 √ 3 Sina = - (4 + 6 √ 3) / √ 3, and C is an acute angle sinc = 1 / 3. According to the relationship between a and C in the normal theorem, we can find

Given the function f (x) = √ 3sinxcosx + cos square X-1 / 2, X ∈ R (1) Find the minimum positive period and monotone increasing interval of function

(1 + cos2x) / 2-1 / 2 = (√ 3 / 2) sin2x + (1 + cos2x) / 2-1 / 2 / 2 = (3 / 2) sin2x + (1 / 2) cos2x = sin2x * cos (π / 6) + cos2x * sin (π / 6) = sin (2X + π / 6) (1) t = 2 π / 2 = 2K π - π / 2 ≤ 2x + π / 6 ≤ 2K π + π + π / 22K π - 2 π - 2 π - π / 6 ≤ 2K π + π / 22K π - 2 π - 2 π - 2 π - 2 π / 6 ≤ 2K π + π - 22K π - 2 π - 2 π - 2 π3 ≤ 2x ≤ 2K π + π / 3

The known function f (x) = cos squared X - √ 3sinxcosx + 1 F (a) =5/6. π /3 ≤ a ≤ 2 π /3 to find the value of sin2a

Because, f (a) = 5 / 6, therefore, sin (π / 6-2a) + 3 / 2 = 5 / 6 = 5 / 6, namely, sin (2A - π / 6 - 2A) + 3 / 6 = 5 / 6, namely, sin (2A - π / 6 - π / 6) = 2 / 3, because, f (a) = 5 / 6, so, sin (π / 6-2a) + 3 / 2 = 5 / 6 = 5 / 6, that is, sin (2A - π / 6) = 2 / 3, because, π / 3 ≤ a ≤ 2 π / 3, so, π / 2 ≤ 2A - π / 6 ≤ 7 π / 6, so, cos (2a2a2a2a2a2a), π / 6 ≤ 7 π / 6 ≤ 7 π / 6, so, cos (2a2a2a2a2a2a2a2aπ - π / 6

Let f (x) = radical 3 sinwx + cos (Wx + π / 3) + cos (Wx - π / 3), w > 0. Find the value range of function f (x). If the minimum positive period of function f (x) is π / 2, then when x belongs to (get to), find the monotone decreasing interval of F (x)

1) (W x + π / 3) + cos (Wx - π / 3) = coswxcos π / 3-sinwxsin π / 3 + coswxcos π / 3 + sinwxsin π / 3 + coswxcos π / 3 + sinwxsin π / 3 = coswxf (x) = root number 3 sinwx + cos (Wx + π / 3) + cos (Wx - π / 3) + cos (Wx - π / 3) = √ 3sinwx + coswx = 2Sin (Wx + π / 6) function f (x) value domain y ∈ [- 2,2] 2) if the function f (x x x x x, x, 2,2,2] 2) if the function f (x x x x (x x x x) is (x) the minimum of

The function f (x) = cos π x The value range of 3 (x ∈ z) is______ .

F (0) = cos0 = 1, f (1) = cos π 3 = 12, f (2) = Cos2 π 3 = 12, f (2) = Cos2 π 3 = - 12, f (3) = cos π = - 1, f (4) = Cos4 π 3 = - 12, f (4) = Cos4 π 3 = 12, f (6) = Cos2 π = 1, f (7) = COS7 π 3 = cos (2 π + π 3) = 12, repeated occurrence, repeated occurrence, f (x) ∈ {1, 12, | 12, | 12, | 12 (x) ∈ {1, 12, | 12, | 12, \ (x) ∈ {1, 12,, | 12, \- - 1}

Let f (x) = sin (ω x + φ), where ω > 0 | φ| π / 2 (1) if cos π / 4cos φ - Si Given the function f (x) = sin (ω x + φ), where ω > 0 | φ | π / 2 (1) if cos π / 4cos φ - SIN3 π / 4sin φ = 0, calculate the value of φ; (2) under the condition of (1), if the distance between two adjacent symmetry axes of the image of function f (x) is equal to π / 3, find the analytic formula of function f (x); and find the minimum positive real number m, so that the function corresponding to the image of function f (x) after M unit length translation to the left is even function

Because the absolute value of ＜ π / 2, so π / 4 + φ is greater than - π / 4 and less than 3 π / 4, so the phase of the image of π / 4 + φ = π / 4 cos π - sin π / 4sin φ = cos π / 4cos φ - sin π / 4sin φ = cos (π / 4 + φ) = 0