Given the function f (x) = 3sin (KX / 5 + π / 3), where k ≠ 0, find the maximum value of the function

Given the function f (x) = 3sin (KX / 5 + π / 3), where k ≠ 0, find the maximum value of the function

kx/5+π/3=2kπ+π/2
When x = 10K π / K + 5 π / 6K (n is a positive integer)
y manx=3

It is known that the function f (x) = 3sin [(K / 5) x + pi / 3] (k > 0, K belongs to Z) has a symmetry axis X = pi / 6, and there is at least one maximum between any two integers
Given that the function f (x) = 3sin [(K / 5) x + pi / 3] (k > 0, K belongs to Z) has an axis of symmetry x = pi / 6, and the maximum and minimum values appear at least once between any two integers, the minimum value of K is obtained
PI is pi

When x = π / 6, f (x) = ± 3
That is, (K / 5) * (π / 6) + π / 3 = 2m π ± π / 2
That is, k = 60m + 5 or K = 60m-25 (m ∈ z)
The distance between the maximum and minimum is at least t / 2

It is known that the function f (x) = 3sin (KX / 5 + π / 3) (k > 0, K ∈ z) has a symmetry axis X = π / 6,
And the maximum and minimum values appear at least once between any two integers

When x = π / 6, f (x) = ± 3
That is, (K / 5) * (π / 6) + π / 3 = 2m π ± π / 2
That is, k = 60m + 5 or K = 60m-25 (m ∈ z)
The distance between the maximum and minimum is at least t / 2