The vector a = (sinwx, Radix 3sinwx) vector b = (sinwx, coswx), w > 0, f (x) = vector a * vector B, and the minimum positive period of F (x) is π, The vector a = (sinwx, Radix 3sinwx) vector b = (sinwx, coswx), w > 0, f (x) = vector a * vector B, and the minimum positive period of F (x) is' π '. (1) find the monotone decreasing interval of F (x), (2) find the value range of F (x) when x belongs to [0,2 π / 3]

(1)f(x)=a·b
=(sinwx)^2+√3sinwxcoswx
=1/2+(√3/2sin2wx-1/2cos2wx)
=1/2+sin(2wx-π/6)
Because t = 2 π / 2W = π, w = 1
So f (x) = 1 / 2 + sin (2x - π / 6)
When π / 2 + 2K π ≤ 2x - π / 6 ≤ 3 π / 2 + 2K π, f (x) decreases
So the decreasing interval is [π / 3 + K π, 5 π / 6 + K π]
(2) Because of π ≤ 3 π
Then - π / 6 ≤ 2x - π / 6 ≤ 7 π / 6
That is - 1 / 2 ≤ sin (2x - π / 6) ≤ 1
So 0 ≤ f (x) ≤ 3 / 2

Given the vector M = (- 1, coswx + Radix 3sinwx), n = (f (x), coswx), where w > 0, If f (3 / 2A + Pai / 2) = (1 + radical 3) / 2, C = radical 3b, calculate the value of a and B

(1) F (x) = 0.5 + sin (2wx + Pai / 6) cycle = 3 Pai, w = 1 / 3
(2) F (3 / 2A + Pai / 2) = (1 + radical 3) / 2 a = 30 degrees
Area = 0.5 bcsina, BC = 4, 3, so B = 2, C = 2, 3
Using cosine theorem to find a = 2

It is known that the minimum positive period of the function f (x) = (Radix 3sinwx + coswx) coswx-1 / 2 (W > 0) is 4 π. Find the monotone increasing interval of F (x)

In this paper, it is shown that (x) = (3 sinwx + coswx) coswx-1 / 1 / 2 = 3 / 3 / 2 (2 SiNW x cosw x) + 1 / 2 (2cos (W x-1) = (3 / 2) sin (2wx) + 1 / 2cos (2wx 0 = sin (2wx + π / 6), 2 π / (2WW)) = 4 π, w = 1 / 2, f (x) = sin (1 / 2), f (x) = sin (1 / 2), f (x) = sin (1 / 2), f (x) = sin (1 / 2) f (x) = sin (1 / 2), f (2WW) = 4 π (2W) = 4 π (W) = w = 1 / 2 (1 / 2) f (x) = f (x) = f (x) = sinsinx + π / 6) when

Let f (x) = coswx (radical 3 * sinwx + coswx), where 0

F (x) = coswx (radical 3 * sinwx + coswx)
=(radical 3) coswxsinwx + (coswx) ^ 2
=[(radical 3) / 2] sin2wx + (cos2wx) / 2 + 1 / 2
=sin(2wx+π/6)+1/2
1. From 2 π / 2W = π, w = 1
So f (x) = sin (2x + π / 6) + 1 / 2
When - π / 6

Given vector M = (1, coswx), vector n = (sinwx, root 3), (W > 0), function f (x) = m * n And the coordinates of the highest point on the f (x) image are (π / 12,2), and the coordinates of the lowest point adjacent to it are (7 π / 12, - 2)

F (x) = sinwx + radical 3 * coswx = 2Sin (Wx + π / 3)
The coordinate of the highest point is (π / 12,2), and the coordinate of the lowest point adjacent to it is (7 π / 12, - 2)
Period = 2 (7 π / 12 - π / 12) = π, so 2 π / w = π, and then w = 2
f(x)=2sin(wx+π/3) =2sin(2x+π/3)

Let the minimum positive period of the function f (x) = sinwx + Radix 3 coswx (W > 0) be π, (1) find the amplitude and initial phase of parallel lines (2) It is shown that the image of function f (x) can be obtained by the exchange of the image of y = SiNx

f(x)=sinwx+√3coswx
=2sin(wx+∏/3)
T=2∏/w=π
W=2
f(x)=2sin(2x+∏/3)
Amplitude A = 2, primary phase Π / 3
(2) Y = SiNx is reduced by 1 / 2 along the X axis to obtain y = sin2x
Then translate Π / 6 to the left along the X axis, and y = sin2 (x + Π / 6) = sin (2x + Π / 3)
Then, it is expanded 2 times along the y-axis to obtain 2Sin (2x + Π / 3)

When the value range of the function y = 2Sin (2x + X / 3) - 1, X ∈ [0, π / 3] is the maximum, the value of X is? Can you be more detailed about that process? I understand

The value range of the function y = 2Sin (2x + X / 3) - 1, X ∈ [0, π / 3] is?

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

π/6<=x<=π/2
π/2<=x+π/3<=5π/6
1/2<=sin（x+π/3)<=1
1<=2sin（x+π/3)<=2
So the value range of the function is [1,2]

Given that f (x) = 2Sin (x + θ / 2) cos (x + θ / 2) + 2 √ 3cos (x + θ / 2) - √ 3, and 0 ≤ θ ≤ π, Let f (x) be an even problem

If f (x) = sin (2x + θ) + 2 √ 3 [1 + cos (2x + θ)] / 2 - √ 3 = sin (2x + θ) + 3cos (2x + θ) = 2Sin (2x + θ + π / 3) is a even function, f (- x) = f (x) 2Sin (- 2x + θ + π / 3) = 2Sin (2x + θ + π / 3) = 2Sin (2x + θ + π / 3) so - 2x + θ + π / 3 = 2K π + 2x + θ + π / 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3, or 3 = 2K π + 2x + 2x + 2x + 2x + π + π + - 2x + θ + π / 3 = 2K π + π - (2x + θ +...)

It is known that f (x) = 2Sin (x + α) 2）cos（x+α 2）+2 3cos2（x+α 2）- 3 is even function and α∈ [0, π] (I) find the value of α; (II) if x is an inner angle of triangle ABC, find the value of X satisfying f (x) = 1

(I) f (x) = 2Sin (x + α 2) cos (x + α 2) + 23cos2 (x + α 2) - 3 = sin (2x + α) + 3cos (2x + α) = 2Sin (2x + α + π 3), if f (x) is an even function, then α + π 3 = k π + π 2, that is, α = k π + π 6, K ∈ Z, ∵ α∈ [0, π].