Given vector M = (COS X / 3, Radix 3cos X / 3) n (SiN x / 3, cosx / 3) function f (x) = m * n Given the vector M = (COS X / 3, Radix 3cos X / 3) n (SiN x / 3, cosx / 3) function f (x) = m * n (1), find the analytic formula of function f (x) (2) find the monotone increasing interval of F (x) (3) if the triangle ABC satisfies B ^ 2 = AC and the angle of edge B is x, try to find the range of X and the range of value of this function f (x)

(1)f(x)=m*n=sin(x/3)cos(x/3)+√3cos²(x/3)
=1/2sin(2x/3)+√3/2cos(2x/3)+√3/2
=sin(2x/3+π/3)+√3/2.
(2) From 2K π - π / 2 ≤ 2x / 3 + π / 3 ≤ 2K π + π / 2 (K ∈ z),
It is found that 3K π - 5 π / 4 ≤ x ≤ 3K π + π / 4 (K ∈ z),
So the monotone increasing interval of F (x) is [3K π - 5 π / 4,3k π + π / 4] (K ∈ z)
(3) Because B 2 = AC,
Therefore, according to the cosine theorem, cosx = (a? + C? - B?) / 2Ac = (a? + C? - AC) / 2Ac ≥ (2Ac AC) / 2Ac = 1 / 2,
If and only if a = C, the equal sign holds
Because x is the inner angle of the triangle, so 0

Given the vector M = (2sinx, cosx), n = (radical 3cosx, 2cosx), the definition function f (x) = loga (m * n-1) (a > 1) 1) Finding the minimum positive period of F (x) 2) Determine the monotone increasing interval of function f (x)

In this paper, the minimum positive period is 2 π / 2 = π when a > 1, when a > 1, then when 2 sin (2x + π / 6 + cos2xsin π / 6 + cos2xsin π / 6) = 2Sin (2x + π / 6) loga (mn-1) = loga [2Sin (2X + π / 6)], so the minimum positive period is 2 π / 2 = π, if a > 1, then when a > 1, then when 2Sin (2x + π / 6) increases monotone monotone increasing, then the 2Sin (2x + π / 6) increases monotonmonotonmonotone increasing, then the minimum positive period is 2 π / 2 = π, when a > 1, then when a > 1, then when 2Sin (2x + π / 6) increases monotone increasing monotone increasing when f (x

Given that vector a = (2sinx, root 3cosx), vector b (cosx, 2cosx), function f (x) = vector a × vector b-1-radical 3, (1) when x ∈ [0, π / 2] Find the maximum value of F (x), that is, the value of X at this time (2) X ∈ R, find the monotone increasing interval of F (x)

f(x)=2sinxcosx+2√3(cosx)^2-1-√3=sin2x+√3cos2x-1=2sin(2x+π/3)-1
(1) When 2x + π / 3 = π / 2, i.e. x = π / 12, the maximum value of F (x) is f (π / 12) = 1
(2)2kπ-π/2

Known vector a＝{2sinx，cosx}， b＝{ 3cosx, 2cosx} define function f (x)= a• b−1． (1) Find the minimum positive period of function f (x) (2) When x ∈ R, find the maximum value of function f (x) and the value of X at this time

F (x) = a * B-1 = 23sinx × cosx + 2cos2x-1 = 3sin2x + cos2x = 2Sin (2x + π 6) (7 points) (1) t = 2 π|ω| = π (9 points) (2) f (x) = 2Sin (2x + π 6)  when 2x + π 6 = π 2 + 2K π (K ∈ z) 

Let f (x) = radical 3 * sin (ω x) + cos (ω x + π / 3) + cos (ω X - π / 3) - 1, (W > 0, X belongs to R), and the minimum positive period of F (x) is π (1) Find the analytic formula of function f (x) and find the minimum value of F (x) (2) In △ ABC, if f (b) = 1, vector Ba * vector BC = (3 root sign 3) / 2, and a + C = 4, find the edge length B

This is not difficult
（1）f(x)=√3sin(ωx)+cos(ωx+π/3)+cos(ωx-π/3)-1=√3sin(ωx)+1/2cos(ωx)-√3/2sin(ωx)+1/2cos(ωx) +√3/2sin(ωx)-1=√3sin(ωx)+cos(ωx)-1=2[√3/2sin(ωx)+1/2cos(ωx)] -1=2sin(ωx+π/3)-1
Because the minimum positive period is π, 2 π / ω = π. Therefore, ω = 2. So f (x) = 2Sin (2x + π / 3) - 1, the minimum value is - 3
(2) F (b) = 1, know that B = π / 12, CoSb = cos (π / 3 - π / 4) = (√ 2 + √ 6) / 4 = [(a + C) ^ 2-2ac - B ^ 2] / 2Ac
Accos (b) = 3 √ 3 / 2. You can calculate the rest by yourself. Typing is too troublesome
That's the way of thinking. The process is basically right
There is also a judgment process when calculating angle B. because the product of two vectors is positive, angle B is an acute angle. You should understand why

Given that the function f (x) = 3 in the root sign is sin (2x minus 6) plus 2Sin squared (x belongs to R), find the minimum positive periodic urgency of function f (x)

F (x) = root 3sin (2x-p / 6) + 2Sin ^ 2 (X-P / 12)
=Root 3sin (2x-p / 6) + 1-cos (2x-p / 6)
=Root 3sin (2x-p / 6) - cos (2x-p / 6) + 1
Therefore, the minimum positive period of F (x) is 2P / 2 = P
Among them: P represents the school

Known function f (x)= 3sin（2x-π 6）+2sin2（x-π 12） (1) Find the minimum positive period of function f (x); (2) Find the set of X when the function f (x) reaches the maximum

(1) F (x) = 3sin (2x - π 6) + 1-cos (2x - π 6) = 2 [32sin (2x - π 6) - 12cos (2x - π 6)] + 1 = 2Sin (2x - π 3) + 1, ∵ ω = 2, ∵ t = π; (2) let 2x - π 3 = 2K π + π 2, K ∈ Z, then x = k π + 5 π 12, K ∈ Z, then the set of X when the function f (x) reaches the maximum value is obtained

Let f (x) = - 2Sin square x + 2 radical sign 3sinxcosx + 1 (1) find the minimum positive period and symmetric center of F (x) (2) If x ∈ [- π in six, π in three], find the maximum and minimum of F (x)

So t = π, the symmetry center is f (x) = 0, 2Sin (x) = 0, so 2x + π / 6 = 0, so 2x + π / 6 = 0, so 2x + π / 6 = 0, so 2x + π / 6 = k π, get x = k π / 2-π / 2-π / 12x ∈ [- π / 6, π / 3], the [- π / 6, π / 6] increases, and [π / 6, π / 3] increases, [π / 6, π / 3] increases, [π / 6, π / 3] increases, [π / 6, π / 3] and [π / 6, π / 3] is [π / 6, π / 3] is [π / 6, π / 3] single minus

The known function f (x) = 2 times the root sign 3sin 2 x-sin (2x - π / 3) Finding the minimum positive period and monotone interval of a function Let α∈ (0, π), f (A / 2) = 1 / 2 + √ 3, and find the value of sin α

(1) Because sin? X = (1-cos2x) / 2, sin (2x - π / 3) = 1 / 2 sin2x - radical 3 / 2 cos2x
So the function f (x) = 2 times the root sign 3sin 2 x-sin (2x - π / 3) = - sin (2x + π / 3) + radical 3
So the minimum positive period of the function is 2 π / 2 = π,
When 2K π - π / 2

Find the monotone increasing interval of the function f (x) = - root 3sin (x - π / 2) + sin (2x - π / 3)

If you can't remember the sum difference product formula, you can find it separately. First, find the monotone decreasing interval of the first one (2k π to 2K π + π), then find the monotone increase of the second (K π - 1 / 12 * π to K π + 5 / 12 * π), and finally find the intersection (it should be 2K π to 2K π + 5 / 6 * π)