It is known that f (x) = 2sinxcosx + 2 root sign 3cos square x-1-radical 3 (2) Find out the symmetry center coordinate and symmetry axis equation of curve f (x) (1) When x belongs to [0. π / 2], find the maximum value of F (x) and the value of X at this time. (3) Find the monotone increasing interval of F (x) on which X belongs to [- 2 / π, 2 / π]. (4) Put the f (x) image to enjoy translation unit after. The resulting image is symmetric about the y-axis. Find the minimum value of M. (5) When x belongs to [0, π], find the x value of F (x) = 0. (6) This paper explains how to get the image of function f (x) from the image of y = SiNx.

X = 2 SiNx cosx + 2 √ 3cos (x) x-1-1-3 = 3 = sin2x + 3 (2cos, x-1) - 1 = sin2x + √ 3cos 2x-1 = 2Sin (2x + π / 3) - 1 from 2x + π / 3 = k π, K ∈ Z: x = k π / 2-π / 6, K ∈ Z, f (x) symmetry center is (K π / 2-π / 6, K ∈ Z, f (x) symmetry center is (K π / 2-π / 6, 6, - 1), K ∈ Z, z from 2x + π / 3 = 3 = 2x + π / 3 = 3 = 3 = 2x + π / 3 = 3 = 1), K ∈ z from K π + π / 2, K ∈ Z

It is known that f (x) = 2 radical 3cos ^ 2 x + SiNx times 2cosx radical 3 Find: 1. The minimum positive period of F (x) 2. Monotone interval of F (x)

∵f（x）=2√3cos² x+2sinxcosx-√3
=√3(1+cos2x)+sin2x-√3
=√3cos2x+sin2x
=2sin(2x+π/3)
(1) The minimum positive period of F (x) t = 2 π / 2 = π
(2) From 2K π - π / 2 = ≤ 2x + π / 3 ≤ 2K π + π / 2,
That is, K π - 5 π / 12 ≤ x ≤ K π + π / 12
From 2K π + π / 2 ≤ 2x + π / 3 ≤ 2K π + 3 π / 2, namely K π + π / 12 ≤ x ≤ K π + 7 π / 12;
The monotonic increase of F (x) is: [K π - 5 π / 12, K π + π / 12]
The monotone decreasing interval is [K π + π / 12, K π + 7 π / 12]; K ∈ Z

The function f (x) = 2cos (2x - π / 6) + 3, X belongs to the range of [- π / 6,2 π / 3]

X belongs to [- π / 6,2 π / 3], and (2x - π / 6) belongs to [π / 6,7 π / 6], so cos (2x - π / 6) belongs to [- 1, √ 3 / 2], f (x) range [1,3 + √ 3]

The range of the function y = 2cos (2x + π / 3) (- π / 6 ≤ x ≤ π / 6)

y=2cos(2x+π/3)
-π/6≤x≤π/6
-π/3≤2x≤π/3
0≤2x+π/3≤2π/3
-1/2≤cos(2x+π/3)≤1
-1≤2cos(2x+π/3)≤2
So the range is [- 1,2]

Why is the value range of function f (x) = 2cos ^ 2x / 2-radical 3sinx [- 1,3]?

F (x) = 2cos ^ 2x / 2-radical 3sinx
=1+cosx-√3sinx
=1+2(1/2*cosx-√3/2*sinx)
=1+2sin(π/6-x)
When sin (π / 6-x) = 1, f (x) max = 3
When sin (π / 6-x) = - 1, f (x) min = - 1

Find the value range of F (x) in [- π / 6, π / 3]

Step 1: power down of angle doubling formula: F (x) = cos2x + 1 + √ 3sin2x step 2: auxiliary angle formula integration: F (x) = √ 3sin2x + cos2x + 1 = 2Sin (2x + π / 6) + 1 because x belongs to [- π / 6, π / 3], we get that: 2x + π / 6 belongs to [- π / 6,5 π / 6], then sin (2x + π / 6) belongs to [- 1 / 2,1], so: F (x) = 2S

Find the value range of the function y = 2cos? X + 2sinx-1

y=2-2sin²x+2sinx-1
=-2(sinx-1/2)²+3/2
-1

The range of the function f (x) = 2sinx-2cos

f（x）=2sinx-2cos
=2（sinx-cos）
=2【√2（1/2sinx-1/2cos）】
=2√2（√2/2sinx-√2/2cos）
=2√2sin（x-π/4）
Because the maximum value of y = sin (x - π / 4) is 1 and the minimum value is - 1
So the value range of the function f (x) = 2sinx-2cos is [- 2 √ 2,2 √ 2]
If it makes sense,

Function y = cos? X + 2sinx (- π / 6 ≤ x ≥ 5 π / 6) range

The solution consists of y = cos? X + 2sinx
=1-sin^2x+2sinx
=-sin^2x+2sinx+1
Let t = SiNx
From - π / 6 ≤ x ≤ 5 π / 6
That is - 1 / 2 ≤ SiNx ≤ 1
That is - 1 / 2 ≤ t ≤ 1
So y = - Sin ^ 2x + 2sinx + 1
It becomes y = - T ^ 2 + 2T + 1 t of [- 1 / 2,1]
So y = - T ^ 2 + 2T + 1
=-（t-1）^2+2
So when t = 1, y has a maximum value of 2
When t = - 1 / 2, y has a minimum of - 1 / 4
Therefore, the value range of the original function is [- 1 / 4,2]

Find the value range of the function f (x) = 2cos square x + 2sinx-1 / 2, X ∈ [- π / 6,5 π / 6]

f（x）=2×（1-sin²x）+2sinx-1/2
=2-2sin²x+2sinx-1/2
Let a = SiNx, - 1 / 2 ≤ a ≤ 1
f=-2a²+2a+3/2=-2（a²-a）+3/2=-2（a-1/2）²+3/2+1/2=-2（a-1/2）²+2
This is a quadratic function. The coefficient of quadratic term is less than 0 and has a maximum value
When x = 1 / 2, f max = 2,
When a = - 1 / 2, Fmin = 0
So the range of F (x) is [0,2]