Find the monotone interval of the function y = 2Sin (2x - sextuple) + 1, the symmetry center and symmetry axis equation of an image

Find the monotone interval of the function y = 2Sin (2x - sextuple) + 1, the symmetry center and symmetry axis equation of an image

Let (2x - sixths) be greater than or equal to (- π / 2 + 2K π) be less than or equal to (π / 2 + 2K π), and then let it be greater than (π X1 / 2 + 2K π) be less than or equal to (π X3 / 2 + 2K π) in this interval

Given the function f (x) = 2Sin (2x - π / 6), X ∈ R. (1) write the symmetric axis equation of function f (x), the coordinates of symmetric center and monotone interval (2) to find the function f (x)（ (2) Find the maximum and minimum values of the function f (x) on the interval [0, π / 6]. Please answer within two days.

If the symmetry axis of F (x) is k π + π / 2, then solve 2x - π / 6 = k π + π / 2, x = (K / 2 + 1 / 3) π, the symmetry center is k π, the coordinate is: ((K / 2 + 1 / 12) π, 0) monotone interval is: (K π - π / 6, K π + π / 3) monotone increase (K π + π / 3, K π + π / 6) monotonically decrease on [0, π / 6]

Given the function y = 2Sin (π x + (π / 6)), find the symmetric axis equation of function y in the interval [21 / 4,23 / 4],

First, write out all the equations of symmetry axis. X = K + 1 / 3 K is an integer. Then solve the inequality 21 / 4 〈 = K + 1 / 3 〈 = 23 / 4, we get k = 5, so the equation of symmetry axis is x = 16 / 3

Find the symmetry axis equation and symmetry center of the image with function y = 1 / 2Sin (2x + π / 6)

2x+π/6=π/2+kπ
2x=π/3+kπ
x=π/6+kπ/2
2x+π/6=kπ
2x=kπ-π/6
x=kπ/2-π/12
The axis of symmetry is x = π / 6 + K π / 2 K ∈ Z
The center of symmetry is (K π / 2 - π / 12,0) k ∈ Z

The function f (x) = log (1 / 2) is a monotone decreasing interval of COS (2x + π / 2)

Firstly, cos (2x + π / 2) is greater than 0,
In addition, cos (2x + π / 2) must be in the increasing range
So 2x + π / 2 is greater than or equal to 2K π - (1 / 2) π and less than or equal to 2K π
It is found that x is greater than or equal to K π - (1 / 2) π and less than or equal to K π - (1 / 4) PI
Timely adoption! Thank you

The function y = cos (2x - π) 3) The monotone decreasing interval of is___ .

∵ for the function y = cos (2x - π)
3) The monotone decreasing interval is 2K π ≤ 2x - π
3≤2kπ+π
That is, K π + π
6≤x≤kπ+2π
Three
So the monotone decreasing interval of function f (x) is [K π + π]
6，kπ+2π
3]（k∈Z）
So the answer is: [K π + π
6，kπ+2π
3]（k∈Z）

What is the monotone decreasing interval of the function y = cos (π / 6) cos (2x) + sin (π / 6) sin (2x)? Is the monotone decreasing interval of COS (π / 6-2x) the same as that of COS (2x - π / 6)?

y=cos(π/6)cos(2x)+sin(π/6)sin(2x)
=cos(2x-π/6)
2X - π / 6 decreases monotonically in [2K π, 2K π + π]
X decreases monotonically in [K π + π / 12, K π + 7 π / 12]

Given the function f (x) = cos (2x - 3) + SiN x - cos x (1), find the monotone decreasing interval of function f (x) (2) if f (a)= (2) If f (a) = three fifths and 2A is the first quadrant angle, find the value of sin 2A

(1)f(x)=1/2cos(2x)+√3/2sin(2x)-cos(2x)=√3/2sin(2x)-1/2cos(2x)=sin(2x-π/6)
The monotone decreasing interval is: 2K π - π / 2 ≤ 2x - π / 6 ≤ 2K π + π / 2K belongs to n
That is, K π - π / 6 ≤ x ≤ K π + π / 3 K belongs to n
(2) The first quadrant angle of F (a) = sin (2A - π / 6) = 3 / 5 > 0 2A - π / 6
2a-π/6=arcsin3/5 2a=arcsin3/5+π/6
sin2a=sin(arcsin3/5+π/6)=sin(arcsin3/5)*cosπ/6+cos(arcsin3/5)*sinπ/6
=3/5*√3/2+4/5*1/2=(3√3+4）/10

Decreasing interval of function cos (5 / 8 π) cos (π / 8-2x) For detailed explanation

1/2π

What is the distance between two adjacent symmetry axes in the image with y = √ 2Sin (2x / 3 + π / 4)?

The distance between two adjacent axes of symmetry is half a period
Obviously, t = 2 π / (2 / 3) = 3 π
Then the half period is 3 π / 2
That is, the distance between two adjacent symmetry axes in the image with y = √ 2Sin (2x / 3 + π / 4) is 3 π / 2