Find the minimum positive period y = 2sinx cosx of the following function

Find the minimum positive period y = 2sinx cosx of the following function

Y = 2sinx cosx = √ 5sin (x - α) (sin α = √ 5 / 5) ‖ the minimum positive period is 2 π

For example, if there is a formula F (x) = (1-cosx) / 2 + 5 / 2sinx+
If there is a formula F (x) = (1-cosx) / 2 + 5 / 2sinx + 1 + cosx, can we multiply by 2?

No, if you multiply by two, then it's twice as big, so that the function is not the same, and only two sides of the equation can be multiplied by two

Four trigonometric functions for senior one
The function y = asin (ω χ + β) (χ∈ R, a > 0, ω > 0, | β) is known|
I need it tonight

(1) According to the coordinates of the highest point, a = 3
The difference between the abscissa of the highest point and the adjacent symmetrical center is 1 / 4
Launch cycle T = pie, launch w = 2
When x = pi / 6, sin (2x + b) = 1
This paper introduces 2x + B = Pie / 2, that is, B = Pie / 6
So y = 3sin (2x + Pie / 6)
(2) The minimum value is sin (2x + Pie / 6) = - 1
That is, when 2x + faction / 6 = k faction + 3 / 2 faction
Introduce x = k school / 2 + 2 school / 3