An axis of symmetry of the image of the function y = sin (3x + π × 3 / 4)... Attention!

The symmetry axis of y = Sint is t = k π + π (K ∈ z)
t=3x+π×3/4.
3x+¾π=kπ+½π（k∈Z）.
3x=kπ-π/4
x=kπ/3-π/12.（k∈Z）
When k = 0,
x=-π/12.

The function f (x) = sin (2 / 3x) cos (2 / 3x) the distance between two adjacent symmetry axes of an image is

f(x)=sin(2/3x)cos(2/3x)=1/2sin(4x/3)
T=2π/(4/3)==3π/2
The distance between the two adjacent symmetry axes is half of the period T, which is 3 π / 4

The symmetry axis of the function y-sin (3x + 5 π / 2)

0

This function can be regarded as f (x) + 1 = sin (3x + π / 6), which is the simplest sine function. When sin (3x + π / 6) is equal to 1 or - 1, X is taken as the axis of symmetry
When x = π / 9, sin (3x + π / 6) = 1

The function f (x) = sin (2x + π / 3) can be expressed as

When 2x + π / 3 = 2K π + π / 2, i.e. x = k π + π / 12 (k is an integer), the maximum value of F (x) is obtained,
When 2x + π / 3 = 2K π - π / 2, that is, x = k π - 5 π / 12 (k is an integer), the minimum value of F (x) is obtained,
Therefore, the symmetry axis equation of the function f (x) = sin (2x + π / 3) image can be
1) X = k π + π / 12 (k is an integer)
2) X = k π - 5 π / 12 (k is an integer)

Find the symmetry center and axis equation of the image with function y = sin (x-pi / 6)

x-pi/6=kπ==>x=kπ+π/6,k∈Z
The symmetry center of image (K π + π / 6,0) k ∈ Z
x-pi/6=kπ+π/2,k∈Z
==>x=kπ+2π/3,k∈Z
The symmetry axis equation of image x = k π + 2 π / 3, K ∈ Z

Let f (x) = √ 3cos ^ 2 ω x + sin ω xcos + a (where ω > 0, a belongs to R) and the abscissa of the first high point of the image of F (x) on the right side of the y-axis is π Let f (x) = √ 3cos ^ 2 ω x + sin ω xcosx + a (where ω > 0, a belongs to R) and the abscissa of the first high point of the image of F (x) on the right side of the y-axis is π / 6 1. Find the value of ω 2. If the minimum value of F (x) on the interval [- π / 3,5 π / 6] is √ 3, find the value of A

1) I have just done this problem, which is reduced to f (x) = cos ^ (2wx-30 ")
So w = 0.5
2) A = radical 3-1

Let f (x) = √ 3 cos 2 ω x + sin ω xcos ω x + A and the abscissa of the first highest point of the image on the right side of the y-axis is π / 6 (1)

f(x)=√3 cos²ωx＋sinωxcosωx＋a
=√3/2(2cos²ωx-1)+1/2＋1/2sin2ωx＋a
=√3/2cos2ωx＋1/2sin2ωx＋a+1/2
=sinπ/3cos2ωx＋cosπ/3sin2ωx+a+1/2
=sin(2ωx+π/3)+a+1/2
When 2 ω x + π / 3 = π / 2, it is the function that takes the first maximum on the right side of the y-axis
That is, 2 * ω * π / 6 + π / 3 = π / 2
We get ω = 1 / 2
Your topic is not complete. According to the existing conditions, you can only do it here

The minimum positive period of the function f (x) = sin 2 (Wx + π / 6) - cos 2 (Wx + π / 6) (W > 0) is 2 π, and the value of W is obtained if TaNx = 4 / 3, And X ∈ (π, 3 π / 2), find the value of F (x)

(x) = sin (Wx + π / 6) - cos (Wx + π / 6) - cos (Wx + π / 6) = - cos [2 (Wx + π / 6)] = - cos (2wx + π / 3) the minimum positive period is 2 π = 2 π / (2W) w = 1 / 2F (x) = - cos (x + π / 3) TaNx = 4 / 3, and X ∈ (π, 3 π / 2) SiNx = - 4 / 5, cosx = - 3 / 5F (x) = - cos (x + π / 3) = cosx, cosx = 3 / 5F (x) = - cos (x + π / 3) = cosx cosx cosx, cosx = 3 / 5F (x) = - cos (x + π / 3) = cosx coss (π

The known function g (x) = 1 X · sin θ + LNX is an increasing function on [1, + ∞), and θ∈ (0, π), f (x) = MX − m − 1 x−lnx (m∈R) (1) Find the value of θ; (2) If f (x) - G (x) is a monotone function in [1, + ∞) function, find the value range of M

(1) The derivation results show that G ′ (x) = - 1sin θ x2 + 1x ≥ 0 holds when x ≥ 1 Ψ 1x ≥ 1sin θ x2 ∵ 1 ≥ 1sin θ · x ∵ θ ∈ (0, π)

An axis of symmetry of the image of the function y = sin (3x + π × 3 / 4)... Attention!