It is known that f (x) = radical 3sin? X + sinxcosx - radical 3 / 2 (x ∈ R) in △ ABC, if a

It is known that f (x) = radical 3sin? X + sinxcosx - radical 3 / 2 (x ∈ R) in △ ABC, if a

F (x) = root sign 3sin? X + sinxcosx radical 3 / 2 = sqrt (3) (1 / 2-cos (2x) / 2) + 1 / 2 * sin2x sqrt (3) / 2=
=1/2*(sin2x-sqrt(3)cos2x)=sqrt(1+3)/2*sin(2x-π/3)=sin(2x-π/3)
f(A)=f(B)=1/2
sin(2A-π/3)=sin(2B-π/3)=1/2
2A-π/3=2B-π/3=π/6
A=B=π/4
Then BC / AB = sin (a) = sqrt (2) / 2

Given the function f (x) = 2cosxsin (x + π / 3) - root sign 3sin ^ 2x + sinxcosx, find the number of equation f (x) = x / 50 π The answer is 399, and the equation of how to dismantle it should also be explained

(x) = 2cosxsin (x + π / 3) - √ 3sin ^ 2x + sinxcosx = cosx (SiNx + √ 3cosx) - √ 3 (SiNx) ^ 2 + sinxcosx = 2ssinxcos x + √ 3 [(cosx) ^ 2 - (SiNx) ^ 2] = sin2x + √ 3cos2x = 2Sin (2X + π / 3) = x / 50 π, sin (2x + π / 3) = x / 100 π, because | sin (2x 2x 2x / 3) = x / 50 π, sin (2x + π / 3) = x / 100 π, because of the | sin (2x 2x 2X / 3) = x / 100 π, because of the | sin (2x 2x 2x+ π / 3)|

The known function f (x) = a• b. Among them a=(2cosx， 3sinx)， b＝(cosx，−2cosx)． (1) Find the function f (x) in the interval [0, π 2] Monotonically increasing interval and range of values on; (2) In △ ABC, a, B and C are the opposite sides of angles a, B and C respectively, f (a) = - 1, and B = 1, and the area of △ ABC is s = 3. Find the value of edge a

（1）f(x)＝2cosx•cosx−2
3sinx•cosx=1−(
3sin2x−cos2x)=1−2sin(2x−π
6) (2 points)
From 2K π + π
2≤2x−π
6≤2kπ+3π
2, K ∈ Z leads to K π + π
3≤x≤kπ+5
6π，k∈Z，
And [0, π
2] The monotonic increasing interval is [π
3，π
2] (4 points)
By − 1
2≤sin(2x−π
6) ≤ 1 ν - 1 ≤ f (x) ≤ 2 ﹤ f (x) ∈ [- 1, 2] (6 points)
（2）∵f（A）=-1，∴A＝π
3, (8 points)
And S = 1
2×1×c×sin600＝
3, ν C = 4 (10 points)
According to the cosine theorem, A2 = B2 + c2-2bccosa = 13A =
13 (12 points)

Given the function f (x) = 3 times sinxcosx + (cosx) ^ 2 (1) Find the minimum positive period of F (x) (2) find the maximum and minimum of F (x) on the interval [- Wu / 6, Wu / 3]

A:
f(x)=√3sinxcosx+(cosx)^2
=(√3/2)*2sinxcosx+(1/2)*[2(cosx)^2-1]+1/2
=(√3/2)sin2x+(1/2)cos2x+1/2
=sin(2x+π/6)+1/2
1) The minimum positive period of F (x) t = 2 π / 2 = π
2）-π/6

The minimum positive period of the function f (x) = sin (2x + ξ) + 2 radical 3cos ^ 2 (x + ξ / 2) - radical 3 is known

F (x) = sin (2x + THR) + 2 radical 3cos ^ 2 (x + thr / 2) - radical 3 = sin (2x + Ψ) + 2 radical 3 [1 + cos (2x + THR)] / 2 - radical 3 = sin (2x + THR) + 3cos (2x + THR) = 2 sin (2x + Ψ + α), so the minimum positive period = 2 π / 2 = π

Given the function f (x) = sin (x / 2) + radical 3cos (x / 2), X belongs to R, find the minimum positive period of F (x), monotone increasing interval on [- 2 π, 2 π]

F (x) = sin (x / 2) + radical 3cos (x / 2)
=2 [1 / 2 * sin (x / 2) + radical 3 / 2 * cos (x / 2)]
=2 sin(x/2+π/3)
The minimum positive period is 4 π
2kπ-π/2≤x/2+π/3≤2kπ+π/2,k∈Z.
4kπ-5π/3≤x≤4kπ+π/3,k∈Z.
Because x ∈ [- 2 π, 2 π],
When k = 0, - 5 π / 3 ≤ x ≤ π / 3
So the monotone increasing interval of the function on [- 2 π, 2 π] is [- 5 π / 3, π / 3]

F (x) = radical (2x-1 / 1-x), if the images of functions y = g (x) and y = f (x) are symmetric about the origin (1) Write the analytic expression of function g (x) (2) Let y = g (x) define the domain as a inequality x ^ 2 - (2a-1) x + a (A-1) ≤ 0. The solution set is B. If a is a proper subset of B, find the value range of A

(1)
f(x)=√[(2x-1)/(1-x)]
(2x-1) / (1-x) ≥ 0, that is (2x-1) / (x-1) ≤ 0
The solution is 1 / 2 ≤ X

It is known that the images of the functions f (x) and G (x) are symmetric about the origin, and f (x) = x2 + 2x (I) find the analytic formula of function g (x); (II) to solve the inequality g (x) ≥ f (x) -| X-1 |

(I) let any point Q (x0, Y0) on the image of function y = f (x) be p (x, y) with respect to the origin point, then p is on the image of G (x), and x0 + x2 = 0y0 + y2 = 0, that is, X0 = - xy0 = - Y. ∵ the point Q (x0, Y0) on the image of function y = f (x), y = - y = x2-2x, that is, y = - x2 + 2x

We know that the images of the functions y = f (x) and y = g (x) are symmetric with respect to the origin, and f (x) = x ^ 2 + 2x (1). Find the analytic formula of the function y = g (x) (2) and solve the inequality G (x) ≥ f (x) -| X-1|

Because f (x) and - f (- x) are symmetric about the origin, G (x) = - f (- x) = - [(- x) ^ 2 + 2 * (- x)] = - x ^ 2 + 2x conclusion: F (x) and f (- x) are symmetric about the Y axis. F (x) and - f (x) are symmetric about the X axis

The image of the function f (x) = sin (2x + φ) is symmetric with respect to the line x = π / 8

The axis of symmetry must pass through the lowest or highest point of the trigonometric function image,
The image of the function f (x) = sin (2x + φ) is symmetric with respect to the line x = π / 8,
When x = π / 8, the function takes the maximum value 1 or the minimum value - 1,
That is sin (π / 4 + φ) = ± 1,
π/4+φ=kπ+π/2,k∈Z.
φ=kπ+π/4,k∈Z.