The symmetry axis equation of the function y = cos (2x + π / 3) is

The symmetry axis equation of the function y = cos (2x + π / 3) is

I'm sure I made a mistake upstairs. Don't listen to him
Let 2x + π / 3 = k π to obtain x = [(3K-1) π] / 6

Why do I calculate the coordinates of the symmetry center of the function y = 3tan (2x + π / 3) as x = k π / 4 - π / 6? Why is it different from the answer What are the coordinates of the symmetry center of the function y = 3tan (2x + π / 3) I calculated x = k π / 4 - π / 6. Why is it different from the answer I'm so depressed

The center of symmetry of y = TaNx is (K π / 2,0), K ∈ Z
Therefore, the question:
2x+π/3=kπ/2
2x=-π/3+kπ/2
Get: x = - π / 6 + K π / 4
Therefore, the center of symmetry is (- π / 6 + K π / 4,0) k ∈ Z
PS: the center of symmetry is a point, not a straight line. It should be written in the form of coordinates
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Find the coordinates of the symmetry center of the image with function y = 3tan (2x + π / 3)?

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All y = 0 are symmetric points
So 0 = Tan (2x + pi / 3)
2X + pi / 3 = k * PI (k is an integer)
x=k*pi/2-pi/6
The coordinates of the symmetrical center point are (k * pi / 2-pi / 6,0), and there are infinite numbers

The center of symmetry of the function y = 3tan (2x + π / 3) is

Let t = 2x + π / 3, from the image and properties of the tangent function:
The symmetry center of the function y = Tan t is each point (K π, 0), and K belongs to Z
That is, 2x + π / 3 = k π
The solution is: x = k π / 2 - π / 6
So the symmetry center of the function y = 3tan (2x + π / 3) is (K π / 2 - π / 6,0), and K belongs to Z

Symmetry center of function y = 3tan (2x + π / 6)

The symmetry center of y = TaNx is (K π, 0), and a set of formulas 2x + π / 6 = k π, x = k π / 2 + (k-1 / 6) π, so the symmetry center coordinate of y = 3tan (2x + π / 6) is (K π / 2 + (k-1 / 6) π, 0)

The center of symmetry of the function y = 3tan (1 / 2x + Wu / 3) is

(-2/3π+2kπ,0)
Because the original centrosymmetry (0,0), 1 / 2x + Wu / 3 = 0, calculate X