The known function f (x) = 3 2+ Two 2sin(2x+π 4)． (1) Find the maximum value and minimum positive period of function f (x); (2) Find the monotone increasing interval of F (x)

(1) Because (x) = function
2+
Two
2sin(2x+π
4) When sin, π
4) When = 1,
Function f (x) takes the maximum value: 3
2+
Two
2, the minimum positive period T = 2 π
2＝π
(2) Function f (x) = 3
2+
Two
2sin(2x+π
4) The increasing range of sin (2x + π) is sin (2x + π)
4) The increasing range of
From 2K π - π
2≤2x+π
4≤2kπ+π
2, K ∈ Z, K π - 3 π
8≤x≤kπ+π
8，
So the monotone increasing interval of F (x) is [K π - 3 π
8，kπ+π
8]，k∈Z

Given the function f (x) = 2cos2x cosx + 2Sin ^ 2x, find f (PAI / 4), find the maximum value of function f (x), and the set of X when f () takes the maximum

1）f(π/4)=2cos(π/2)-cos(π/4)+2[sin(π/4)]^2
=0-√2/2+2(√2/2)^2
=1-(√2/2)
2）f(x)=2cos2x-cosx+2(sinx)^2
=2[2(cosx)^2-1]-cosx+2[1-(cosx)^2]
=4(cosx)^2-2-cosx+2-2(cosx)^2
=2(cosx)^2-cosx
=2(cosx-1/4)^2-1/8
Therefore, when cosx = - 1, f (x) takes the maximum value, and f (x) max = 2 + 1 = 3
In this case, x = 2K π + π (K ∈ z)

When 0 is less than X and less than or equal to Pai / 4, find the value of function f (x) = 1 + cos2x + 8sin ^ 2 x / sin2x Is the maximum value of the function f (x) = 1 + cos2x + 8sin ^ 2 x / sin2x cosx / SiNx when 0 is less than X and less than or equal to Pai / 4

F (x) = [1 + cos2x + 8sin? 2x] / [sin2x] = [2cos? X + 8sin? 2x] / [2sinxcosx] = (4tanx) + (1 / TaNx), where 0

How to simplify the known function f (x) = (sin2x + cos2x) 2 - 2Sin? 2x Find the minimum positive period

f(x)=1+2sin2xcos2x-2sin²2x=1+sin4x-(1-cos4x)=sin4x+cos4x=√2sin(4x+π/4)
The minimum positive period T = 2 π / 4 = π / 2

The symmetric axis of the image of the function FX = 2cos (Wx + π / 4) (W > 0) is exactly the same as that of the image of the function GX = 2Sin (2x + α) + 1

The symmetric axis of the image of the function FX = 2cos (Wx + π / 4) (W > 0) is exactly the same as that of the function GX = 2Sin (2x + α) + 1

It is known that the symmetry axes of images with F (x) = 3sin (Wx - π / 6) (W > 0) and G (x) = 2cos (2x + φ) + 1 are identical, If x belongs to [0, π / 2], the value range of F (x) is (),

It is known that the symmetry axes of images with F (x) = 3sin (Wx - π / 6) (W > 0) and G (x) = 2cos (2x + φ) + 1 are identical,
The cycle is the same,
∴w=2
That is, f (x) = 3sin (2x - π / 6)
When x ∈ [0, π / 2]
Then 2x - π / 6 ∈ [- π / 6,5 π / 6]
∴ sin(2x-π/6)∈[-1/2,1]
That is, f (x) ∈ [- 3 / 2,3]

The symmetric axes of the images of the functions f (x) = 3sin (wx-6) and G (x) = 2cos (2x + P) + 1 are exactly the same. What does this mean? Their periods are the same The symmetry axes of the images of the functions f (x) = 3sin (wx-6) and G (x) = 2cos (2x + P) + 1 are exactly the same. What does this mean? Their periods are the same, why?

After reading your answers, I have some different opinions
The symmetry axes of the images of the functions f (x) = 3sin (wx-6) and G (x) = 2cos (2x + P) + 1 are exactly the same. What does this mean? Their periods are the same, why?
Analysis: ∵ the symmetry axes of images with F (x) = 3sin (Wx - π / 6) and G (x) = 2cos (2x + P) + 1 are identical
This shows that the two functions have the same period, that's all
∵g(x)=2cos(2x+p)+1==>T=2π/2=π
∴f(x)=3sin(2x-π/6)
When the periods of the two functions are the same and the symmetry of the images is the same, their phases can be the same or half a period different
 by shifting f (x) = 3sin (2x - π / 6) left half a period, we can get that f (x) = 3sin (2 (x + π / 2) - π / 6) = 3sin (2x + 5 π / 6) = 3cos (2x + π / 3)
By F (x) = 3sin (2x - π / 6) = 3sin (2x + π / 2-2 π / 3) = 3cos (2x-2 π / 3)
/ / g (x) = 2cos (2x + π / 3) + 1 or G (x) = 2cos (2x-2 π / 3) + 1
That is, P = π / 3 or P = - 2 π / 3

The known function f (x) = 3sin (ω X - π) 6) If x ∈ [0, π), the symmetry axes of the images with (ω > 0) and G (x) = 2cos (2x + φ) + 1 are identical 2] Then the value range of F (x) is () A. [-3 2，3] B. （-3 2，3） C. [-3 2，+∞） D. （-∞，3）

From the meaning of the question, we can get that ω = 2, ∵ x ∈ [0, π
2]，∴ωx-π
6=2x-π
6∈[-π
6，5π
6]，
According to the trigonometric function image:
The minimum value of F (x) is 3sin (- π)
6）=-3
The maximum value is 3 sin π
2=3，
So the value range of F (x) is [- 3
2，3]，
Therefore, a

The known function f (x) = 3sin (ω X - π) 6) If x ∈ [0, π), the symmetry axes of the images with (ω > 0) and G (x) = 2cos (2x + φ) + 1 are identical 2] Then the value range of F (x) is () A. [-3 2，3] B. （-3 2，3） C. [-3 2，+∞） D. （-∞，3）

The function f (x) = 3sin (ω x − π) is known 6) (ω > 0) and G (x) = 2cos (2x + π) 3) If the symmetry axes of the two images are identical, then the value of ω is______ .

The symmetric axis equation of the function g (x) = 2cos (2x + π 3) is: 2x + π 3 = k π K ∈ Z, i.e. x = k π 2 − π 6 K ∈ Z, and the symmetric axis equation of function f (x) = 3sin (ω x − π 6) (ω > 0) is: x = k π ω + 2 π 3 ω K ∈ Z, because the functions f (x) = 3sin (ω x − 6) (ω > 0) and G (x

The known function f (x) = 3 2+ Two 2sin(2x+π 4)． (1) Find the maximum value and minimum positive period of function f (x); (2) Find the monotone increasing interval of F (x)