Given a vector a = (2coswx, 1), B = (Radix 3sinwx coswx, n), where x ∈ R, w > 0, the function f (x) = a * B (x ∈ R), if the minimum positive period of F (x) is π., the maximum value is 3 (1) Find the maximum value of function f (x) on X ∈ [0, π / 2] (2) The three angles of △ ABC correspond to three sides of a, B, C, and satisfy f (a) = 2, a = 1, B + C = radical 3 + 1, and calculate the area of △ ABC

(1) F (x) = Radix 3sin2wx-cos2wx + n-1 = 2Sin (2wx - π / 6) + n-1 because t = π, w = 1 because the maximum value is 3, so n = 2, so f (x) = 2Sin (2x - π / 6) + 1, so the minimum value of function f (x) on X ∈ [0, π / 2] is 1 and the maximum value is 3 (2) f (a), so a = 2 / 3

Given that the period of the function FX = Radix 3sinwx * coswx cos ^ 2wx (W > 0) is a two-part derivation, the value of W and the monotone increasing interval of the function are obtained

The period T = 2 π / | w | = π, then w = 1; at this time, f (x) = sin (2x X - π / 6) - (1 / 2). The periodic period T = 2 π / |w | w | = π, then w = 1; at this time, f (x) = sin (2x - π / 6) - (1 / 2) increasing interval is 2K π - π / 2 ≤ 2x π / 6 ≤ 2K π / 6 ≤ 2K π + π / 2, we get: K π π / 6 ≤ x ≤ K π + π / 3 / 3: K π - π / 6 ≤ x ≤ K π + π + π / 3 / 3: K π - π - π - πit's a good idea

Let f (x) = radical 3 * (coswx) ^ 2 + (sinwx) * (coswx) + a (where w > 0, a belongs to R) Let f (x) = √ 3 (coswx) ^ 2 + sinwxcoswx + a (where w > 0, a belongs to R), and the abscissa of the first lowest point of the image of F (x) on the right side of the y-axis is 7 Pai / 6. (1) find the value of W (2) if the minimum value of F (x) on the interval [- Pai / 3,5 Pai / 6] is √ 3, find the value of A There must be a process

The abscissa of the first lowest point of F (x) = √ 3 (coswx) ^ 2 + sinwxcoswx + a = radical 3 (cos2wx + 1) / 2 + sin2wx / 2 + a = sin (2wx + π / 3) + √ 3 / 2 + A, the abscissa of the first lowest point on the right side of the y-axis is 7 π / 6

1. 2. Function y = 3sin (2x - π / 3), find the maximum sum of Y 1. Y = 1 / 2 sin (4x - π / 3) 2. The function y = 3sin (2x - π / 3) is used to find the maximum and minimum value of Y, the minimum positive period, frequency, phase, monotone increasing interval, monotone decreasing interval and symmetry center Thank you for your kindness~ The shortest time to reply to the bonus points

If x is any real number, the range is - 1 / 2

The function y = 2Sin (2x - π / 3) x belongs to the range of [π / 3,3 π / 4]

Let t = 2x - π / 3, then π / 3 < = T < = 7 π / 6. Y = 2sint
It is easy to see that the maximum value is obtained at t = π / 2, ymax = 2; the minimum value is obtained at t = 7 π / 6, and Ymin = 2Sin (7 π / 6) = 2 * (- 1 / 2) = - 1. [in fact, the function increases in [π / 3, π / 2] and decreases in [π / 2,7 π / 6]
So: the range is [- 1,2]

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

x∈(-π,-π/2)
x/3∈(-π/3,-π/6)
x/3+π/3∈(0,π/6)
y∈(0,1/2)

Find the value range of the function y = 3-2sin (x + π / 6), X ∈ [0,2 π] thankou!

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From - π 2 + 2K π ≤ 2x + π 3 ≤ π 2 + 2K π, we get − 5 π 12 + K π ≤ x ≤ π 12 + K π, i.e. the increasing range of the function is [- 5 π 12 + K π, π 12 + K π], K ∈ Z, where 2x + π 3 = π 2 + 2K π, i.e. x = π 12 + K π, K ∈ Z, i.e., the symmetric axis of the function is x = π 12 +

Given that f (x) = 2Sin (2wx + π / 6), if the straight line x = π / 3 is a symmetric axis of the image of function f (x), find W

The symmetry axis of F (x) = 2Sin (2wx + π / 6) is known
2wx+π/6=kπ+π/2
x=kπ/(2w)+π/6w
If the line x = π / 3 is a symmetric axis of the image of function f (x),
kπ/(2w)+π/6w=π/3 k∈Z
w=(3k+1)/2 k∈Z

Let f (x) = 2Sin (2x - π / 3) + 1. (1) find the monotone interval and symmetry axis of function f (x); (2) Let g (x) = a · f ((1 / 2) x + π / 6) - 2Sin 2 x + 1. (a ∈ R), find the expression H (a) of the maximum value of function g (x)

（1）
Increasing interval
2kπ-π/2≤2x-π/3≤2kπ+π/2
That is 2K π - π / 6 ≤ 2x ≤ 2K π + 5 π / 6
The increasing interval is [K π - π / 12, K π + 5 π / 12]
The interval π / 12 π + 11 is the same
Axis of symmetry,
2x-π/3=kπ+π/2
∴ x=kπ/2+5π/12
（2）
g(x)=a*(2sinx+1)-2sin²x+1
=-2sin²x+2asinx+a+1
Let SiNx = t
y=-2t²+2at+a+1,-1≤t≤1
The symmetry axis t = A / 2, and the image opening is downward
① A / 2 ≤ - 1, that is, a ≤ - 2
When t = - 1, y has the maximum value - A-1
② -1