If a symmetry axis of F 'is a straight line x = π 4, then a possible value of θ is () A. 512πB. -512πC. 1112πD. -1112π

If a symmetry axis of F 'is a straight line x = π 4, then a possible value of θ is () A. 512πB. -512πC. 1112πD. -1112π

The analytic expression of image f obtained by translation is y = 3sin (x - θ - π 3) + 3. The equation of symmetry axis is X - θ - π 3 = k π + π 2 (K ∈ z). Substituting x = π 4 into θ = - 7 π 12-k π = (- k-1) π + 5 π 12 (K ∈ z), making k = - 1, θ = 512 π, so a is selected
Given 2x + x ^ 2Y = 2, find the value of - 3x ^ 2y-6x + 7
2x+x^2y=2
Multiply by (- 3)
-6x-3x^2y=-6
So - 3x ^ 2y-6x + 7 = - 6 + 7 = 1
Multiply left and right by - 3 to get - 3x ^ 2y-6x = - 6, so the original formula is equal to 1
x^2y=2-2x,
-3x^2y-6x+7=-3*(2-2x)-6x+7=1
From 2x + x ^ 2Y = 2 to x ^ 2Y = 2-2x
-3x^2y-6x+7
=-3（2-2x)-6x+7
=-6+6x-6x+7
=1
-3x^2y-6x+7=-3*（2x+x^2y）+7=-3*2+7=1
If the function f (x) = sin (ω x + π 6) (ω > 0), the image of function f (x) is shifted to the right by π 6 unit length, and the equation of symmetry axis of the image is x = π 3, then the minimum value of ω is______ .
The image of the function f (x) = sin (ω x + π 6) (ω > 0) is shifted to the right by π 6 unit lengths. The analytic expression of the function corresponding to the image is y = sin [ω (x - π 6) + π 6] = sin (ω x + π 6 - ω π 6). The equation of the axis of symmetry is x = π 3, ω· π 3 + π 6 - ω π 6 = k π + π 2, K ∈ Z, that is, ω π 6
Given 3x-2y = 5, find the value of the algebraic formula [6x ^ 4 + (x ^ 3-y) ^ 2 - (x ^ 3 + y) ^ 2] / (- 2x) ^ 3
【6x^4+(x^3-y)^2-(x^3+y)^2】÷(-2x)^3=[6x^4+x^6-2x³y+y²-x^6-2x³y-y²]÷﹙-8x³﹚=(6x^4-4x³y﹚÷﹙-8x³﹚=-3/4x+1/2y=-1/4(3x-2y)=1/4×5=-5/4
Original formula = (6x ^ 4 + x ^ 6-2x & # 179; y + Y & # 178; - x ^ 6-2x & # 179; Y-Y & # 178;) / (- 8x & # 179;)
=(6x^4-4x³y)÷(-8x³)
=-(3x-2y)/4
=-5/4
Finding monotone interval of function y = sin (tt / 2x + tt / 3)
Monotone interval of y = sin (π / 2 * x + π / 3)
2kπ-π/2≤π/2* x+π/3≤2kπ+π/2,k∈Z
2k-1/2≤1/2* x+1 /3≤2k+1/2,k∈Z
Find the simple increasing interval of X
2kπ+π/2≤π/2* x+π/3≤2kπ+3π/2,k∈Z
2k+1/2≤1/2* x+1 /3≤2k+3/2,k∈Z
Find the simple minus interval of X
-2X ^ 2 + 10x-5 ≤ 0, (3x + 1) (x-4) (x-3) > 0, 1 / 6 of x > 1 / 3x, find the inequality of each X
1、x≤10-2√5/4,x≥10+2√5/4
2. (- 1 / 30 or X
x≤10-2√5/4,x≥10+2√5/4
The monotone increasing interval of the function y = sin (π / 3-x) is
y=sin(π／3-x ）=-sin(x-π/3)
To find the increasing interval of the original function is to find the decreasing interval of y = sin (x - π / 3)
∴ 2kπ+π/2≤x-π/3≤2kπ+3π/2
That is 2K π + 5 π / 6 ≤ x ≤ 2K π + 11 π / 6
The monotone increasing interval is [2K π + 5 π / 6,2k π + 11 π / 6], K ∈ Z
Solving inequality system (1 / 3) (X-6)
(1/3)(X-6)
Finding monotone increasing interval of function y = sin (1-x) π / 2
This kind of problem can benefit all one's life by mastering the method. Sin (1-x) π / 2 can be transformed into - sin (x-1) π / 2, which requires the monotone increasing interval of sin (1-x) π / 2, that is, the monotone increasing interval of - sin (x-1) π / 2, that is, the monotone decreasing interval of sin (x-1) π / 2, Let sin (x-1) π / 2 be greater than or equal to 2K π + π / 2 and less than or equal to 2K π + 3 π / 2, (k belongs to Z) solve this inequality
Hope you can master this method, because it is very useful in high school mathematics!
[2K, 2K + 1], K takes all integers
Solving equations or inequalities: (1) 3 (2x-1) (x + 6) - 5 (x-3) (x + 6) = x (x + 9) (2) (2x-1) & sup2; + (1-2x) (1 + 2) x ≥ 0
(1) 3(2x--1)(x+6)--5(x--3)(x+6)=x(x+9)
（x+6)(6x--3--5x+15)=x^2+9x
(x+6)(x+12)=x^2+9x
x^2+18x+72=x^2+9x
9x=--72
x=--8.
(2) (2x--1)^2+(1--2x)(1+2x)>=0
(2x--1)(2x--1--1--2x)>=0
(2x--1)(--2)>=0
2x--1>=0
x>=1\2.