The function f (x) = sin ^ 2wx + √ 3 * coswx * cos (π / 2-wx) (W > 0) is known, and the distance between two adjacent symmetry axes of the image with function y = f (x) is π / 2 (1) Find the symmetry center of the image of function y = f (x) (2) When x ∈ [0, π / 2] and f (x) = a has a real number solution, the value range of real number a is obtained

Given the function f (x) = sin ^ 2wx + √ 3 * coswx * cos (π / 2-wx) (W > 0), and the distance between two adjacent symmetry axes of the image of function y = f (x) is π / 2 (1) find the symmetric center of the image of function y = f (x). (2) when x ∈ [0, π / 2], and f (x) = a has a real number solution, find the value range of real number a

The known function f (x) = √ 3sin (Wx + φ) - cos (Wx + φ) (0

f(x)=2sin(wx+φ-π/6)
φ - π / 6 = k π, so φ = k π + π / 6
Take k = 0 and φ = π / 6
T = 2 π / w = π, so w = 2
f(x)=2sin(2x)

Let f (x) = 2A cos ^ 2 x + B SiN x cos X - radical 3 / 2, and f (0) = radical 3 / 2, f (Wu / 4) = 1 / 2 1. Find the minimum positive period of a function 2. Find the monotone decreasing interval of function

By F (0) = √ 3 / 2, f (π / 4) = 1 / 2, there is a = √ 3 / 2, B = 1
Thus f (x) = √ 3cos ^ 2 x + SiN x cos X - √ 3 / 2
=√3cos^2 x+（1/2）sin2x-√3/2
=(√3/2)cos2x+（1/2）sin2x
=sin(2x+π/3)
1,T=π
2,2k π + π / 2 ≤ 2x + π / 3 ≤ 2K π + 3 π / 2

Let f (x) = radical 3 * cos ^ 2 * Wx + sinwxcoswx + a (where w > 0, alpha belongs to R), and the abscissa of the first high point of the image of F (x) on the right side of the Y axis is Wu / 6 (1) Find the value of W (2) If the minimum value of F (x) on the interval [- Wu / 3,5 μ / 6] is the radical 3, find the value of A (1) Find the value of W (2) If the minimum value of F (x) on the interval [- Wu / 3,5 μ / 6] is the radical 3, find the value of A

W = 1 / 2 a = (root three of two) + 1

Let f (x) = radical 3 / 2-radix 3sin ^ 2wx sinwxcoswx (W > 0) and y = f (x), the distance from one symmetry center to the nearest symmetry axis is pi / 4. (1) find the value of W (2) find the maximum and minimum value of F (x) on the interval [Pai, 3pai / 2] (for detailed problem-solving process, please attach the image picture explanation, Don't write it by yourself,

Analysis: (1) f (x) = V 3/2- V 3sin^2 ω x-1/2*sin2 ω x = V 3/2- Ⅴ 3/2* (1-cos2 ω x) -1/2*sin2 ω x = V 3/2*cos2 ω x-1/2*sin2 ω x = -sin (2 ω x- π /3) because y = one of the images of F (x)

The function f (x) = 2coswx (Radix 3sinwx + coswx) is known, where w > 0, and the distance between the symmetry axes of two adjacent lines of the image of function f (x) is π 1: If f (x) = 2, find cos ((2 π) / 3-x)? 2: in the triangle ABC, the opposite sides of angle ABC are a, B, C respectively, and satisfy (2a-c) CoSb = bcosc?

The function f (x) = 2coswx (root: 3sinwx + coswx), where w > 0 = 2 √ 3sinwx + 2 (coswx) ^ 2 = √ 3sin2wx + cos2wx + 1 = 2Sin (2wx + π / 6) + 1 is known, the period T = 2 π, so, w = 1 / 2, f (x) = 2Sin (x + π / 6) + 1 = 2Sin (x + π / 6) + 1 = 2Sin (x + π / 6) + 1 = 2Sin (x + π / 6) = 1 / 2, cos (x + π / 6) = 1 / 2, cos (x + π / 6) = 1 / 2, cos (x + π / 6) = 1 / 2, cos (x + π / 6) = 1 / 2, cos (/ 2, (no corner

The vector a = (1 + coswx, 1), B = (1, a + radical 3sinwx), f (x) = AB has a maximum value of 2 on R 1. Find the value of real number a 2. If y = GX is an increasing function on (0, Pai / 4), the maximum value of W can be obtained

f(x)=ab
=1+coswx+a+√3sinwx
=a+1+2sin(wx+π/6)
(1) The maximum value of F (x) on R is 2
a+1+2=2
a=-1
f(x)=2sin(wx+π/6)
(2)y=g(x)=2sinwx
Y = g (x) is an increasing function on (0, π / 4)
g(0)=0,0

When w is greater than 0, the vector M = (1,2coswx), n = (Radix 3sin2wx, - coswx), f (x) = Mn, the distance between two symmetrical axes of adjacent images is pi / 21 to find W 2 find the maximum and minimum value of [Pai / 4, Pai / 2]

Vector M = (1,2cos ω x), n = (√ 3sin2 ω x, - cos ω x) f (x) f (x) = Mn = √ 3sin2 ω X-2 (COS ω x) ^ 2 = √ 3sin2 ω x-cos2 ω X-2 (COS ω x) ^ 2 = √ 3sin2 ω x-cos2 ω X-1 = 2Sin (2 ω x-π / 6) - 1 image adjacent 2 symmetry axis distance π / 2 explains that t / 2 = π / 2, so t = π / 2, so t = π / 2 ω = π, then ω = 12. Know from 1 that f (x) = 2Sin (x) = 2Sin (2x (2x 2x 2x 2x 2x 2x 2x 2x (2x) - π

Given that the vector a = (radical 3, - 1) B = (sinwx, coswx) w > 0, f (x) = a * B, and the minimum positive period of F (x) is a card, find w? When 0 < x < π / 2, what is the range of F (x)?

f(x)=a*b＝V3 Sin wx - 1 Cos wx=2(V3/2 Sinwx - 1/2 Cos wx)
=2[SinwxCos(pi/6)- CoswxSin(pi/6)]
=2Sin(wx-pi/6)

The vector a = (Radix 3sinwx, coswx) B = (coswx, coswx) w > 0 f (x) = a * b f (x) the minimum positive period is π ① Find w 2 when 0

f(x)＝a*b＝√3sinwx·coswx＋coswx·coswx＝sin(wx＋π/6)＋1/2,
(1) So (f) is the smallest period;
(2) When 0