The interval (2x sin / 2 π) is monotone Finding monotone interval by y = 2Sin (2x - π / 6)

The interval (2x sin / 2 π) is monotone Finding monotone interval by y = 2Sin (2x - π / 6)

The monotone increasing interval is 2K π - π / 2 ≤ 2x - π / 6 ≤ 2K π + π / 2K π - π / 6 ≤ x ≤ K π + π / 3
The monotone decreasing interval is 2K π + π / 2 ≤ 2x - π / 6 ≤ 2K π + 3 π / 2K π + π / 3 ≤ x ≤ K π + 5 π / 6

Finding monotone interval by y = 2Sin (2x + π / 6) - 1 And the maximum and the set of X when the maximum value is reached

The maximum value is 1
When sin (2x + π / 6) = 1
In this case, 2x + π / 6 = π / 2 + 2K π
x=π/6+kπ
The minimum value is - 3
When sin (2x + π / 6) = - 1
In this case, 2x + π / 6 = - π / 2 + 2K π
x=-π/3+kπ

How to find the monotone interval of 2Sin (- 2x + π / 6)

The coefficient before x is converted to a positive number, y = 2Sin (- 2x + π / 6) = - 2Sin (2x - π / 6) (1) increase interval, that is, the decrease interval of y = 2Sin (2x - π / 6), 2 = 2K π + π / 2 ≤ 2x - π / 6 ≤ 2K π + 3 π / 2,2,2k π + 3 π / 2,2k π + 2 π + 2 π + 2 π / 3 ≤ 2x ≤ 2K π + 5 π / 3 (K π + 3 ≤ x ≤ K π + 5 π / 6), namely the increase interval is [K π + K π + 5 π + 5 π / 6, that is [K π + K π + K π + 5 π + 5 π / 3, K

The monotone increasing interval of y = 2Sin (2x - π / 6) on [0, π]

Let - π / 2 + 2K π ≤ 2x - π / 6 ≤ π / 2 + 2K π, K ∈ Z
That is - π / 6 + K π ≤ x ≤ π / 3 + K π, K ∈ Z
∵x∈[0,π]
If k = 0, then - π / 6 ≤ x ≤ π / 3
Let k = 1, then there is 5 π / 6 ≤ x ≤ 4 π / 3
The monotonic increasing interval of the function on [0, π] is [0, π / 3] u [5 π / 6, π]

The maximum value of the function f (x) = 2Sin (x + 6 / π) + A on [- 5 / 12 π, π, 12] is

-5π/12≤x≤π/12
π/6-5π/12≤x+6/π≤π/6+π/12
-π/4≤x+6/π≤π/4
-Radical 2 / 2 ≤ sin (x + 6 / π) ≤ Radix 2 / 2
-Radical 2 + a ≤ 2Sin (x + 6 / π) + a ≤ radical 2 + A
The maximum value of the number f (x) = 2Sin (x + 6 / π) + A on [- 5 / 12 π, π, 12] is root 2 + a

Y = 2Sin (2x + π / 6) get the function y = g (x) and find the maximum value of the function y = g (x) in the interval [0, π / 12]

Because 0 ≤ x ≤ π / 12
So π / 6 ≤ 2x + π / 6 ≤ π / 3
Then the maximum value of y = 2Sin (2x + π / 6) on [0, π / 12] is 2Sin (π / 3) = √ 3
The maximum value of y = g (x) on [0, π / 12] is √ 3

Sum of maximum and minimum values of function y = 2Sin (π X / 6 - π / 3) (0 ≤ x ≤ 9))

The original function y = 2Sin (π X / 6 - π / 3), because 0 ≤ x ≤ 9, the value range of π X / 6 - π / 3 is as follows
-π/3≤πx/6-π/3≤7π/6
When π X / 6 - π / 3 = π / 2, that is, x = 5, the function takes the maximum value, that is, the maximum value Y (max) = 2
Due to the limitation of the value range of X, we can't get the ideal minimum value
When x = 0, π X / 6 - π / 3 = - π / 3. The minimum value of the function is y (min) = - 2 * 3 ^ (1 / 2) / 2 = - 3 ^ (1 / 2)
Therefore, the sum of the maximum and minimum of the function: Y (max) + y (min) = 2 + [- 3 ^ (1 / 2)] = 2-3 ^ (1 / 2)

The known function f (x) = 2Sin (π - x) cosx (1) Find the minimum positive period of F (x); (2) Find the monotone increasing interval of F (x) and its symmetry axis

∵f（x）=2sin（π-x）cosx=2sinxcosx=sin2x
(1) According to the periodic formula, t = π can be obtained
(2) Order − 1
2π+2kπ≤2x≤1
2 π + 2K π, − π
4+kπ≤x≤π
4+kπ，k∈Z
Let 2x = k π + 1
2 π, x = π
4+kπ
2，k∈Z
The monotonic increasing range of F (x) is [− π
4+kπ，π
4 + K π], its axis of symmetry x = π
4+kπ
2，k∈Z

The known function f (x) = 2Sin x cosx-2sinx squared (x ﹤ R) ① Find the minimum positive period; 2. Find the value range of function on [π / 24. π / 4] The process should be detailed and urgent

F (x) = 2Sin x cosx-2sinx square = sin2x - (1-cos2x) = sin2x + cos2x-1 = √ 2Sin (2x + π / 4) - 1, so the period of the period = 2 π / 2 = 2 π, 2 = 2 π, 2 = π, X ∈ [π / 24. π / 4] 2x + π / 4 ∈ [π / 3,3 π / 4] x = π / 4, the minimum value is the minimum value when the minimum value = 2 × √ 2 / 2-1 = 0x = π / 8, when the maximum value is the maximum value = √ 2Sin (2 2 2 2 2 2 / 2-1 = 0 = π / 8, the maximum value of the maximum value = √ 2Sin (2 sin2 (2 x π / 8

Given the square · X of the function f (x) = (SiNx + cosx) ^ 2-2sin, find the minimum positive period of the function f (x)

F (x) = (SiNx + cosx) ^ 2-2sin square · x
=1+2sinxcosx-2sin^2x
=cos2x+sin2x
So we can see from the above: Tmin = 2pai / 2
The minimum positive period of F (x) is: Tmin = Pai