The image of y = sin (2x-3 π) can be obtained by shifting π / 6 to the left I'm reviewing how to calculate the left translation π / 6,

The image of y = sin (2x-3 π) can be obtained by shifting π / 6 to the left I'm reviewing how to calculate the left translation π / 6,

Y = sin2x --- shift π / 6 to the left to get: sin2 (x + π / 6) = sin (2x + π / 3)

Given the function y = sin2x + 2sinxcosx + 3cos2x, X ∈ R, then (I) what is the minimum positive period of the function? (II) on what interval is a function increasing?

（Ⅰ） y=sin2x+2sinxcosx+3cos2x
=（sin2x+cos2x）+sin2x+2cos2x
=1+sin2x+（1+cos2x）
=Sin2x + cos2x + 2
=
2sin(2x+π
4) + 2, - 5 points
The minimum positive period of the function is π
(II) by 2K π - π
2≤2x+π
4≤2kπ+π
2, K ∈ Z -- (8 points)
K π - 3 π is obtained
8≤x≤kπ+π
8
The increasing interval of the function is: [K π − 3 π
8，kπ+π
8] , K ∈ Z -------- (12 points)

Find the minimum value of the function y = sin2x + 2sinxcosx + 3cos2x, and write the set of X that makes the function y take the minimum value

y=sin2x+2sinxcosx+3cos2x
=（sin2x+cos2x）+2sinxcosx+2cos2x
=1+sin2x+（1+cos2x）
=2+sin2x+cos2x
=2+
2sin（2x+π
4）．
When sin (2x + π)
4) When = - 1, y gets the minimum value of 2-
If and only if 2x + π
4=2kπ-π
2 is x = k π - 3
When 8 π, take the minimum,
The set of minimum x is {x | x = k π - 3
8π，k∈Z}．

Write the set of the most important function sin ^ 2 What formula is used?

y=sin^2x+2sinxcosx+3cos^2x
=sin^2x+2sinxcosx+cos^2x+2cos^2x
=1+sin2x+2cos^2x-1+1
=sin2x+cos2x+2
=√2(√2/2sin2x+√2/2cos2x)+2
=√2(sin2xcosπ/4+cos2xsinπ/4)+2
=√2sin(2x+π/4)+2
The maximum value of Y is √ 2 + 2
x=kπ+π/8

The known function y = sin2x + 2sinxcosx + 3cos2x, X ∈ R (1) Find the monotone increasing interval of the function; (2) Find the maximum value of the function and the corresponding value of X; (3) The equation of symmetry axis and the coordinate of symmetry center of the function are obtained

y=sin2x+2sinxcosx+3cos2x=1-cos2x
2+sin2x+3(1+cos2x)
Two
=sin2x+cos2x+2=
2sin(2x+π
4) + 2. (5 points)
(1) By - π
2+2kπ≤2x+π
4≤π
2 + 2K π to - 3 π
8+kπ≤x≤π
8+kπ(k∈Z)．
So the monotone increasing interval of the function is [- 3 π
8+kπ，  π
8 + K π] (K ∈ z). (8 points)
(2) Let 2x + π
4=π
2 + 2K π, x = π
8+kπ(k∈Z)，
So when x = π
When 8 + K π (K ∈ z), ymax = 2+
2. (12 points)
(3) By 2x + π
4=π
2 + K π, x = π
8+kπ
2(k∈Z)，
So the equation of symmetry axis of this function is x = π
8+kπ
2(k∈Z)．
By 2x + π
4 = k π, x = - π
8+kπ
2(k∈Z)，
Therefore, the symmetry center of the function is: (- π)
8+kπ
2, 2) (K ∈ z). (16 points)

To get the image of the function y = sin (2x + 1 / 2), how to translate all the points on the image of the function y = sin2x

1 / 4 units to the left
Written as sin2 (x + 1 / 4)
Left plus right minus
You'll know

To get the image of the function y = sin (2x + quarter π), simply shift the graph of y = sin2x to the left by at least () units

Y = sin (2x + 4 parts π) = sin2 (x + 8 parts π)
To get the image of the function y = sin (2x + quarter π), simply shift the graph of y = sin2x to the left by at least (π / 8) units

How to translate the image of function y = sin2x to get the image of function y = sin (2x + π / 4)

Shift right 8 / π
A little bit forgetful. It should be

In order to get the image of function y = 2 + sin (2x + π / 6), we only need to translate the image vector of function y = sin2x

In order to get the image of the function y = 2 + sin (2x + π / 6), we only need to shift the image vector (- π / 12,2) of the function y = sin2x

After the image of function y = sin (2x + π / 6) is translated by vector a, the function corresponding to the new image is y = sin2x, then the vector a = () A （π/6,0） B （-π/6,0） C （π/12,0） D （-π/12,0）

C