The minimum period of the function y = 2cos (x / 2 - π / 6) is?

The minimum period of the function y = 2cos (x / 2 - π / 6) is?

Tmin = 2 π / 1 / 2 = 4 π

The minimum positive period of the function y = 1 / 2cos - 1 is How to write the formula.

y=1/2cos²x-1
=1/2(2cos²x-2)
=1/2(2cos²x-1)-1/2
=1/2cos2x-1/2
Minimum positive period = 2 π / 2 = π

What is the minimum positive period of the function y = 2cos (π / 2 - 2x)?

Y = 2Sin 2x, t = pie

The known function FX = 2cos ^ 2 (X-6) + 2Sin (x-4) cos (x-4) - 1 1 to find the minimum positive period of function FX and the symmetric axis equation of image 2. Find the value range of function FX on the interval [- 12, 2]

Because f (x) = 1 + cos (2x - π / 3) + sin (2x - π / 2) - 1 = 1 / 2cos2x + radical 3 / 2sin2x-cos2x = radical 3 / 2sin2x-1 / 2cos2x = sin (2x - π / 6), the minimum positive period T = 2 π / 2 = π, the symmetry axis equation: 2x - π / 6 = k π + π / 2, that is: x = 1 / 2K π + π / 3 (k belongs to Z) because - π / 12

The function FX = √ 3sin (2x Pai / 6) - 2cos 2 (x-pai / 12) + 1 and X ∈ R is known (1) Finding the maximum and minimum positive period of function FX (2) The function of function g (x) is obtained by shifting the image of function FX to the left by six units. The monotone increasing interval of function g (x) and the equation of symmetry axis are obtained

f(x) = √3sin(2x-π/6)-2cos²(x-π/12) + 1
= √3sin(2x-π/6)-cos(2x-π/6)
= 2{sin(2x-π/6)cosπ/6-cos(2x-π/6)sinπ/6}
= 2sin(2x-π/6-π/6)
= 2sin(2x-π/3) ∈【-2,2】
Minimum value, - 2, maximum value 2, minimum positive period 2 π / 2 = π
By shifting the image of function f (x) to the left π / 6 units, the function g (x) is obtained
g(x) = 2sin[2(x+π/6)-π/3] = 2sin2x
When 2x ∈ (2k π - π / 2,2k π + π / 2), G (x) increases monotonically
So g (x) monotonically increasing interval x ∈ (K π - π / 4, K π + π / 4)
Symmetry axis equation x = k π / 2 ± π / 4

The maximum and minimum positive period of the function y = 2Sin (2x - π / 6)

The maximum and minimum positive period of the number y = 2Sin (2x - π / 6)
Max = 2 x 1 = 2
Minimum positive period = 2 π △ 2 = π

The minimum positive period of the function y = 2Sin (2x + π / 6) + 1

PI

The minimum positive period of the function y = 2Sin (1 / 2x + π / 6) is the answer. Write it clearly. Thank you

The minimum positive period of sin itself is 2 π,
+π / 6 doesn't affect it, it just shifts the image to the right
1 / 2x multiply the period by 2, because the image has been stretched twice, and now the increment of X is half the original

The period of the function y = 2Sin (1 / 2x - π / 6) is

T=π/W
w=1/2
T=4π

Find the function y = sin (π 3+4x）+cos（4x-π 6) The period, monotone interval and the maximum and minimum of

∵（π
3+4x）+（π
6-4x）=π
2，
∴cos（4x-π
6）=cos（π
6-4x）=sin（π
3+4x），
The original formula is y = 2Sin (4x + π)
3) The minimum positive period of this function is 2 π
4, i.e. t = π
2．
When - π
2+2kπ≤4x+π
3≤π
When 2 + 2K π (K ∈ z), the function increases monotonically, so the monotonic increasing interval of the function is [- 5 π]
24+kπ
2，π
24+kπ
2]（k∈Z）．
When π
2+2kπ≤4x+π
3≤3π
When 2 + 2K π (K ∈ z), the function decreases monotonically, so the monotone decreasing interval of the function is [π]
24+kπ
2，7π
24+kπ
2]（k∈Z）．
When x = π
24+kπ
2 (K ∈ z), ymax = 2;
When x = - 5 π
24+kπ
2 (K ∈ z), Ymin = - 2