The symmetry axis of function y = 1 / 2Sin (x - π / 3)

The symmetry axis of function y = 1 / 2Sin (x - π / 3)

So dy / DX = 0y = (1 / 2) sin (x - π / 3) dy / DX = (1 / 2) cos (x - π / 3) if dy / DX = 00 = (1 / 2) cos (x - π / 3) = 0x - π / 3 = π / 2x = 5 π / 6 axis of symmetry, x = 5 π / 6

The symmetric axis of the function y = 2Sin (2x - π / 3) is

X=5π/12+kπ/2

The function y = 2Sin (2x + π / 3) has an axis of symmetry and a center of symmetry A.x=π/12,（-π/6.,0） B.x=π/6,（π/12,0） C.x=π/3,(π/3,0） D.x=-π/12,（5/6π,0）

Axis of symmetry:
2x+π/3=kπ+π/2
x=kπ/2+π/12,k=0,x=π/12
Centrosymmetry
2x+π/3=kπ+π
x=kπ/2-2π/3 k=1,x=-π/6
So choose a

Function y = 2Sin (x / 3 - π / 6) of the image of a symmetric axis, to find a straight line

The axis of symmetry of SiNx is where SiNx takes the maximum value
That is, x = k π + π / 2
So x / 3 - π / 6 = k π + π / 2
So x = 3K π + 2 π

If the symmetric axis of the function y = 2Sin (x + θ) (where θ belongs to (0, π / 2)) is x = π / 3, then θ is?

x+θ=π/2
π/3+θ=π/2
θ=π/6

The (π) is a symmetric line (π = 2x, π = 2x)

The symmetry axis of ∵ y = 2Sin (2x + α) is a straight line x = π / 8
∴①2sin(2*π/8+α)=2
2sin(π/4+α)=2
∴π/4+α=π/2+2kπ
α=π/4+2kπ
∵-π

Find the value range of the function y = 2cos ^ x + 5sinx-4

y=2cos²x+5sinx-4
y=2(1-sin^2x)+5sinx-4
=-2sin^2x+5sinx-2
=-2(sinx-5/4)^2+9/8
range
When SiNx = 1, there is a maximum of y = 1
When SiNx = - 1, there is a minimum of y = - 72 / 8 = - 9
Range [- 9,1]

It is known that the function y = root 2cos square X - root 2Sin square x + 2 root sign 2sinxcosx + 1, X belongs to R, (1) find the value range and minimum positive period of the function, (2) find the value range of the independent variable x when the function goes to the maximum value

solution
y=√2cos²x-√2sin²x+2√2sinxcosx+1
=√2(cos²x-sin²x)+√2sin2x+1
=√2sin2x+√2cos2x+1
=2(√2/2sin2x+√2/2cos2x)+1
=2(sin2xcosπ/4+cos2xsinπ/4)+1
=2sin(2x+π/4)+1
∵sin(2x+π/4)∈[-1,1]
∴-1

Find the value range of the function y = 2cos? X + 5sinx-4 (π / 3 ≤ x ≤ 5 π / 6)

Let f (x) = 2cos? X + 5sinx-4
Then f (x) = 2 (1-sin? X) + 5sinx-4
=-2sin²x+5sinx-2
=-2(sinx-5/4)²+9/8
It is known from (π / 3 ≤ x ≤ 5 π / 6)
1/2≤sinx≤1
So f (x) min = f (1 / 2) = 0
f(x)max=f(1)=1
So the range f (x) ∈ [0,1]

Find the value range of the function y = 2Sin ^ 2x-3sinx + 1

Let SiNx = t, where t ∈ [- 1,1]
y=2t^2-3t+1=2(t-3/4)^2-1/8
When t = 3 / 4, the minimum value y = - 1 / 8
When t = - 1, the maximum value y = 6
Range [- 1 / 8,6]
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