Given the function f (x) = 2Sin (x + α / 2) * cos (x + α / 2) + 2 √ 3cos ^ 2 (x + α / 2) -√ 3, α is a constant. If 0 is less than or equal to α, less than or equal to 180 degrees Let f (x) be the α value of even function

Given the function f (x) = 2Sin (x + α / 2) * cos (x + α / 2) + 2 √ 3cos ^ 2 (x + α / 2) -√ 3, α is a constant. If 0 is less than or equal to α, less than or equal to 180 degrees Let f (x) be the α value of even function

F (x) can be reduced to = = = sin (2x + a) + 3cos (2x + a). Then it can be converted into the same name, that is = = 2Sin (2x + A + π / 3). In order to make the sine function even, a + π / 3 = π / 2 + K π. And because a is less than π, a is equal to π / 6,

Given the function f (x) = 2Sin (x / 2 - π / 3) + 1, find the symmetry axis and center of function y = f (x)

x/2-π/3=kπ＋π/2
x/2=kπ＋5π/6
Namely
The symmetry axis is: x = 2K π + 5 π / 3
Center of symmetry:
The abscissa is satisfied
x/2-π/3=kπ
x/2=kπ＋π/3
x=2kπ＋π/3
The ordinate is 1
therefore
The symmetry center is (2k π + π / 3,1)

The function y = 2Sin (x + π) 3) An axis of symmetry of an image is () A. x＝−π Two B. x=0 C. x＝π Six D. x＝−π Six

Let x + π
3=π
2 + K π (K ∈ z), x = π
6+kπ（k∈Z），
The function y = 2Sin (x + π)
3) The symmetry axis equation of the image is x = π
6+kπ（k∈Z），
Take the integer k = 0 to get x = π
6 is a symmetry axis of the function image
Therefore, C

The known function f (x) = 1 + 2Sin (2 ω x + π) 6) If the line x = π 3 is an axis of symmetry of the graph of function f (x) (1) Find ω and the minimum positive period (2) Find the monotone decreasing interval of function f (x), X ∈ [- π, π]

(1) It can be seen from the problem that: 2 ω· π 3 + π 6 = k π + π 2 (K ∈ z), so there is ω = 12 + 32K (3) f (x) = 1 + 2Sin (x + π 6), thus the period of the function is t = 2 π If π 2 + 2K π ≤ x + π 6 ≤ 3 π 2 + 2K π, π 3 + 2K π can be obtained

Find the value range of the following functions: (1) y = 2Sin (x + 3 parts of π), X belongs to [6 parts of π, 2 parts of π] (2) y = 2cos square x + 5sinx-4

The first one is: because x belongs to the school of 6 / 2 to the school of 2 / 2, so the one in brackets belongs to the school of 2 / 6 to the school of 5 / 6. Through the sine image, we can see that the maximum value of Y at the 2 / 2 School is the minimum value of the 5 / 6 school, so the value range is [2, 1】 Second, we need to change the element. First, we can convert the square of cosine into the square of 1-sine, which is y = 2-2 times of sine square + 5 times of sine-4. The simplification is y = - 2 sine square plus 5 times sine-2. Next, we regard sine as a variable t, which is equivalent to a parabola. Note that the range of T is [- 1, 1】 The axis of symmetry is 5 / 4, so when t is equal to - 1, the minimum value is taken when t is equal to 1, and the maximum value is taken when it is equal to 1

Find the value range of the function y = (2cos ^ 2x + 1) (2Sin ^ 2x + 1)

y=(2cos^2x+1)(2sin^2x+1)
=4sin²xcos²x+2sin²x+2cos²x+1
=(2sinxcosx)²+3
=sin²(2x)+3
Sin2x ∈ [- 1,1]
Then sin 2 (2x) ∈ [0,1]
sin²(2x)+3∈[3,4]
Range = [3,4]

Given the function y = 2Sin (3x + π / 3), X belongs to R When x belongs to [- 2 π / 9, π / 6], find the maximum and minimum value of the function

x∈[-2π/9,π/6]
3x+π/3∈[-π/3,5π/6]
sin(3x+π/3)∈[-√3/2,1]
2sin(3x+π/3)∈[-√3,2]
Maximum value of function = 2
Minimum value of function = - √ 3

Given the function f (x) = 2Sin (x - π / 3) + 1, if the period of function y = f (KX) (k > 0) is 2 π / 3, when x ∈ [0, π / 3], the equation f (KX) = m has exactly two different solutions?

F (KX) = 2Sin (KX - π / 3) + 1 first period T = 2 π / 3 because t = 2 π / k = 2 π / 3, so k = 3 because x ∈ [0, π / 3] so KX ∈ [0, π]
Let KX = n, so f (n) = 2Sin (n - π / 3) + 1, u = n - π / 3 ∈ [- π / 3,2 π / 3] have two different solutions, that is, y = m has two intersections with it. According to the function image, m ∈ [radical 3 + 1,3] can be seen

Known function y = 1 / 2Sin (1 / 2x + π / 3) Find out the definition field, value range, minimum positive period, monotone increasing interval, and what is the maximum value of y when x is taken,

Definition domain: C
Range: C
Period: 4 π
The function is unbounded

The function f (x) = 3sin (2x - π) 6) In the interval [0, π 2] The value range on is______ .

∵x∈[0，π
2]，
∴2x∈[0，π]，
∴2x-π
6∈[-π
6，5π
6]，
∴sin（2x-π
6）∈[-1
2，1]，
∴f（x）=3sin（2x-π
6）∈[-3
2，3]；
That is, f (x) is in the interval [0, π]
2] The value range on is [- 3
2，3]．
So the answer is: [- 3
2，3]．