Let f (x) = cos2x + 2 The maximum value of 3sinxcosx (x ∈ R) is m and the minimum positive period is t (I) find m and t; (II) if there are 10 unequal positive numbers Xi satisfying f (XI) = m and Xi < 10 π (I = 1,2 X 1 + x 2 + +The value of X10

∵f(x)＝
3sin2x+cos2x＝2sin(2x+π
6) (4 points)
（Ⅰ）∵M=2
∴T=2π
2 = π (6 points)
(II) ∵ f (XI) = 2, that is, 2Sin (2xi + π)
6)＝2
∴2xi+π
6＝2kπ+π
2，
∴xi＝kπ+π
6 (K ∈ z) (9 points)
And 0 ＜ Xi ＜ 10 π, νk = 0, 1 , 9 (11 points)
∴x1+x2+… +x10＝(1+2+… +9)π+10×π
6=140
3 π (12 points)

The maximum value of the function f (x) = Radix 3sin2x + 2cos ^ 2x + m on the interval [0, π / 2] is 6, Then the center of symmetry of F (x) is

F (x) = √ 3 sin2x + 2 (cosx) ^ 2 + M = √ 3 sin2x + cos2x + m + 1 = 2Sin (2x + π / 6) + (M + 1) because f (x) max = 6, M = 3, from 2x + π / 6 = k π, x = k π / 2 - π / 12, so the center of symmetry of F (x) is (K π / 2 - π / 12,4) k ∈ Z [

The maximum value of a function f (x) = Radix 3sin2x + 2cos? X = m on the interval [0,2 parts π] is 6 (1) To find the value of Changshu m and the symmetry center of F (x) image (2) The graph of function F1 (x) is obtained by making the symmetric axis image of function f (x) with respect to the Y axis, Then the image of function F1 (x) is shifted to the right by 4 / π units to get the image of function F2 (x), Finding monotone decreasing interval of function F2 (x)

1,f(x)=√3sin2x+2cos²x+m
=√3sin2x+1+cos2x+m
=2sin(2x+π/6)+m+1
∵0≤x≤π/2 ∴π/6≤2x+π/6≤7π/6
Then f (x) max = 2 + m + 1 = 6, M = 3
Let 2x + π / 6 = (2k + 1) π, then x = k π + 5 π / 12
Then the center of symmetry is (K π + 5 π / 12,4) (K ∈ z)
2. If the point (x, y) is on F1 (x), then the point (- x, y) is on f (x)
Then y = F1 (x) = f (- x) = 2Sin (- 2x + π / 6) + 4 = - 2Sin (2x - π / 6) + 4
Then F2 (x) = - 2Sin [2 (x - π / 4) - π / 6] + 4 = - 2Sin (2x-2 π / 3) + 4
Let 2K π - π / 2 ≤ 2x-2 π / 3 ≤ 2K π + π / 2
kπ+π/12≤x≤kπ+7π/12
Then the monotone decreasing interval of function F2 (x) is [K π + π / 12, K π + 7 π / 12] (K ∈ z)

Let the minimum value of the function y = 2cos 2x + root sign 3sin2x + a (a is a real constant) and then the minimum value on the interval [0, π / 2] is - 4, then the value of a is equal to

Y=2cos²x＋√3sin2x+a
=1+cos2x＋√3sin2x+a
=2（1/2cos2x＋√3/2sin2x）+1+a
=2sin（2x+π/6）+1+a
When x = 0, sin (2x + π / 6) has a minimum value of 1 / 2
2*1/2+1+a=-4
a=-6
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Given the function f (x) = 2cos? X + root sign 3sin2x + A, find the maximum and minimum value of function f (x) when x ∈ [0, π / 2] Given the function f (x) = 2cos? X + root sign 3sin2x + A, the maximum and minimum values of function f (x) when x ∈ [0, π / 2] are obtained Given the function FX = sin (2x + π / 6) + cos (2x + π / 3) (1) find the minimum positive period of function f (x). (2) find the minimum value of function f (x) and the set of X when f (x) takes the minimum value. (3) find the range of X values for y ≤ 0

The solution f (x) = 2cos? X + radical 3sin2x + A
=2cos²x-1+√3sin2x+a+1
=cos2x+√3sin2x+a+1
=2（√3/2sin2x+1/2cos2x）+a+1
=2sin（2x+π/6)+a+1
By X ∈ [0, π / 2]
Know 2x ∈ [0, π]
That is 2x + π / 6 ∈ [π / 6,7 π / 6]
That is - 1 / 2 ≤ sin (2x + π / 6) ≤ 1
That is - 1 ≤ 2Sin (2x + π / 6) ≤ 2
That is, a ≤ 2Sin (2x + π / 6) + A + 1 ≤ a + 3
That is, a ≤ f (x) ≤ a + 3
Therefore, the maximum value of F (x) is a + 3 and the minimum value is a
2 f（x）=sin(2x+π/6)+cos(2x+π/3)
=sin(2x+π/6)+sin[π/2-(2x+π/3)]
=sin(2x+π/6)+sin(π/6-2x）
=sin(2x+π/6)-sin(2x-π/6）
=sin2xcosπ/6+cos2xsinπ/6-[sin2xcosπ/6-cos2xsinπ/6]
=2cos2xsinπ/6
=cos2x
So the period of the function T = 2 π / 2 = π
2 when 2x = 2K π + π, K belongs to Z, y has the minimum value - 1
In other words, when x = k π + π / 2, K belongs to Z, y has the minimum value - 1
Therefore, the minimum value of function y = f (x) is - 1, and the set of corresponding x is {X / x = k π + π / 2, and K belongs to Z}
3 by Y ≤ 0
That is, cos2x ≤ 0
In other words, 2K π + π / 2 ≤ 2x ≤ 2K π + 3 π / 2, K belongs to Z,
That is, K π + π / 4 ≤ x ≤ K π + 3 π / 4, K belongs to Z
Therefore, the value range of X of Y ≤ 0 {X / K π + π / 4 ≤ x ≤ K π + 3 π / 4, K belongs to Z}

Let f (x) = sin (x + π / 3) + 2 radical sign 3sin ^ 2 x / 2 (1) find the maximum value of F (x) and X at this time (2) the opposite side lengths of the inner angles a, B and C of △ ABC are a, B and C respectively, if f (b) = √ 3, B = √ 6 / 2, C = 1

In this case, the x-π / 3 = 1 / 2sinx + 3 / 2cosx + √ 3 (1-cosx) 3 (1-cosx) = 1 / 2 SiNx - √ 3 / 2cosx + √ 3 = sin (x-π / 3) + 3 when sin (x-π / 3) = 1, Fmax = 1 + √ 3, x-π / 3 = 2K π + π / 2 x = 2K π + 5 π / 6 K / 6 K ∈ Z f (b) = 3f (b) = sin (b-π / 3) + √ 3sin (B - B-3) + 3 (b-sin (b-π / 3) + 3sin (b-b-3) B-3 (B-3) + 3sin (B-3) B-3, B π / 3) = 0, B = π / 3B = √ 6 / 2, C = 1

Given a vector = (SiNx, root 3 / 4), B vector = (COS (x + π / 3), 1), the function FX = a vector times b vector (1) The maximum and monotone decreasing interval of FX (2) It is known that in the triangle ABC, the opposite sides of the angle ABC are ABC. F (a) = 0 and a = radical 3, respectively

F (x) = vector A. vector b
=sinxcos(x+π/3)+√3/4.
=(1/2)[sin(x+x+π/3)+sin(x-(x+π/3)]+√3/4.
=(1/2)[sin(2x+π/3)-sinπ/3]+√3/4
=(1/2)[sin(2x+π/3)-√3/2]+√3/4
=(1/2)sin(2x+π/3)-√3/4+√3/4.
∴f(x)=(1/2)sin(2x+π/3).
(1) When sin (2x + π / 3) = 1, that is, 2x + π / 3 = π / 2. X = π / 12, f (x) has a maximum value and f (x) max = (1 / 2)
When sin (2x + π / 3) = - 1, i.e. 2x + π / 3 = 3 π / 2, x = 7 π / 12, f (x) has a minimum value and f (x) min = - (1 / 2)
The monotone decreasing interval of ∵ SiNx is 2K π + π / 2

Let f (x) = a vector b vector, where a vector = (SiNx, - radical 3 / 2) B vector = (COS (x + 3 π), - 1 / 2), X belongs to R (1) Finding the maximum value and monotone increasing interval of function f (x) (2) If the image of function f (x) is translated along the x-axis so that the translated image becomes a centrosymmetric figure with respect to the coordinate origin, how can the translation distance be minimized?

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Given vector a = (SiNx, - 1), B = (root 3cosx, - 1 / 2), function f (x) = (vector a + vector b) * A-2 1. Find the minimum positive period of function f (x) 2. We know that a, B, C are the opposite sides of a, B and C, where a is an acute angle, a = 2 root sign 3, C = 4, and f (a) = 1, find the area s of a, B and △ ABC

Vector a = (SiNx, - 1), vector b = ((√ 3) cosx, - 1 / 2), function f (x) = (a + b) · A-2;
It is known that a, B and C are the opposite sides of a, B and C, respectively, where a is an acute angle, a = 2 √ 3, C = 4, and f (a) = 1,
Find the area s of a, B and triangle ABC
One
a+b=(sinx+(√3)cosx,-1-1/2)=(sinx+(√3)cosx,-3/2)；
So f (x) = (a + b) · A-2 = [SiNx + (√ 3) cosx] SiNx + 3 / 2-2 = sin? 2x + (√ 3) sinxcosx-1 / 2
=(1-cos2x)/2+(√3/2)sin2x-1/2=(√3/2)sin2x-(1/2)cos2x=sin2xcos(π/6)-cos2xsin(π/6)
=sin(2x-π/6)
Because f (a) = sin (2A - π / 6) = 1,
Therefore, 2A - π / 6 = π / 2, 2A = π / 2 + π / 6 = 2 π / 3,
∴A=π/3.
Two
From the cosine theorem, there is a  2 = B  2 + C  2 - 2bccosa, substituting the known value 12 = B  2 + 16-4b, that is, B  4B + 4 = (b-2) 2 = 0, so B = 2;
SΔABC=(1/2)bcsinA=(1/2)×2×4×sin(π/3)=2√3.

The vector a = (COS ω x, sin ω x, vector b = (COS ω x, Radix 3cos ω x) where (0

(1)
F (x) = cos ω x * coswx + Radix 3sinwx * cosx
=1 / 2cos2wx + 1 / 2 + radical 3 / 2 * sin2wx-1 / 2
=sin(2wx+π/6)
One axis of symmetry is π / 6, so:
2W * π / 6 + π / 6 = π / 2 + K π (k belongs to integer 0

Let f (x) = cos2x + 2 The maximum value of 3sinxcosx (x ∈ R) is m and the minimum positive period is t (I) find m and t; (II) if there are 10 unequal positive numbers Xi satisfying f (XI) = m and Xi < 10 π (I = 1,2 X 1 + x 2 + +The value of X10