Given that the opposite sides of the inner angles a, B, C in the acute triangle ABC are a, B, C, the vector m is defined=

From m ∥ n, 2sinb, [(2cos? B / 2) - 1] = (- radical 3) cos2b
That is, 2sinbcosb = - root 3cos2b
And ∵ B is an acute angle,

Given SiNx + cosx / 2sinx-3cosx = 3, find the value of 1, Tan (π - x), 2, 5 / 2Sin? X-3cos? X, 3. The value of 2Sin? X-3sin (3 π + x) cos (π - x) - 3cos? X

First of all, the numerator and denominator of TaNx are divided by cosxtanx + 1 / 2tanx-3 = 3, and then the value of TaNx is 21

Given the function f (x) = 2Sin 2 (π / 4 + x) + √ 3cos2x-1, X ∈ R (1), find the minimum positive period and monotone increasing interval of function f (x) (2) In the triangle ABC, if f (c) = √ 3,2sinb = cos (A-C) - cos (a + C), find the value of Tana

f(x) = 2sin²(π/4+x)+√3cos2x-1
= {1-cos[2(π/4+x)] + √3cos2x - 1
= -cos(π/2+2x) + √3cos2x
= -sin2x + √3cos2x
= -2(sin2xcosπ/3 - cos2xsinπ/3)
= -2sin(2x-π/3)
Minimum positive period = 2 π / 2 = π
2X - π / 3 ∈ (2k π + π / 2,2k π + 3 π / 2), where k ∈ Z increases monotonically
Monotonically increasing interval: (K π + 5 π / 12, K π + 11 π / 12), where k ∈ z-2sin (2x - π / 3)
C∈（0,π）
2C-π/3∈（-π/3,5π/3）
f(C)=√3
-2sin(2x-π/3)=√3
sin(2x-π/3)=-√3/2
2C-π/3=4π/3
C=5π/6
B=π-(A+C)
sinB=sin(A+C)
2sinB=cos(A-C)-cos(A+C)
2sin(A+C)=cos(A-C)-cos(A+C)
2sinAcosC+2cosAsinC = cosAcosC+sinAsinC-(cosAcosC-sinAsinC)
2sinAcosC+2cosAsinC = 2sinAsinC
-√3sinA+cosA= sinA
(√3+1)sinA=cosA
tanA = 1/(√3+1) = (√3-1)/2

A symmetric axis equation of the function f (x) = Radix 3sin (x + π / 6) + cos (x + π / 6) is

f(x)=2*(√3/2*sin(x+π/6)+1/2*cos(x+π/6))=2sin(x+π/6+π/6)=2sin(x+π/3) ,
The axis of symmetry satisfies x + π / 3 = π / 2 + K π,
Therefore, the equation of symmetry axis is x = π / 6 + K π, K ∈ Z
Let k = 0 to obtain a symmetric axis equation x = π / 6

The known function f (x) = 2 √ 3sinxcosx-3sin? X-cos? X + 2 If the opposite sides of the inner angles a, B, C of the triangle ABC are a B C and B / a = √ 3, [sin (2a + C)] / Sina = 2 + 2cos (a + C), find the value of F (b)

f(x)=2√3sinxcosx-3sin²x-cos²x+2=√3(2sinxcosx)-(sin²x+cos²x)-(2sin²x+1) +3=√3sin(2x)-1-cos(2x)+3=√3sin(2x)-cos(2x) +2=2[(√3/2)sin(2x)-(1/2)cos(2x)]+2=2sin(2x-π/6)+2b/a=√...

The known function f (x) = 3sin? X + 2 √ 3sinxcosx + cos? X.x ∈ R

f(x)=(√3sinx+cosx)²
=4(√3/2 * sinx+cosx*1/2)²
=4(sinx*cos(π/6)+cosx*sin(π/6))²
=4sin²(x+π/6)

Function y = 3sin2x-2 The value range of 3sinxcosx + 5cos2x is______ .

Function y = 3
2（1-cos2x）-
3sin2x+5
2（1+cos2x）=cos2x-
3sin2x+4=2cos（2x+π
3）+4，
∵-1≤cos（2x+π
3）≤1，
∴2≤2cos（2x+π
3）+4≤6，
Then the value range of function y is [2,6]
So the answer is: [2,6]

It is known that the minimum positive period of the function f (x) = √ 3sin ω xcos ω x-cos ^ 2 ω x + 1 / 2 (ω > 0, X ∈ R) is π / 2 1. The value of F (2 π / 3) and the coordinates of the symmetry center of the image of the function f (x) are obtained 2. When x ∈ [π / 3, π / 2], find the monotone increasing interval of function f (x) Note: π = Pi

(x) f (x) = √ 3sin ω x-cos (x-cos) x-cos (2) x + 1 / 2 = 3 / 2 * (2 SiNW x coswx) - cos ^ 2wx + 1 / 2 = √ 3 / 2sin2wx-cos ^ 2wx + 1 / 2 = √ 3 / 2sin2wx - [(1 + cos2wx) / 2] + 1 / 2 = √ 3 / 2sin2wx-1 / 2cos2wx = - cos (π / 3 + 2wx) because the minimum positive period is π / 2, so t = 2 = 2, t = 2 (π / 3 + 2wx) because the minimum positive period is π / 2, so t = 2 = 2, because the minimum positive period is π / 2, so t = 2, so t = 2, because the minimum positive period is π / 2W = π / 2, w = 2

Given vector a = (SiNx, - 1), vector b = (root 3cosx, - 1 / 2), function f (x) = vector a + vector b) * vector A-2 (1) Find the value range of function f (x); (2) given that a, B, C are the opposite sides of a, B, C of triangle ABC, a = 2, root 3, and f (a) = 1, find the maximum area of a and triangle ABC

Vector a = (SiNx, - 1), vector b = ((√ 3) cosx, - 1 / 2), function f (x) = (a + b) · A-2;
It is known that a, B and C are the opposite sides of a, B and C, respectively, where a is an acute angle, a = 2 √ 3, C = 4, and f (a) = 1,
Find the area s of a, B and triangle ABC
a+b=(sinx+(√3)cosx,-1-1/2)=(sinx+(√3)cosx,-3/2)；
So f (x) = (a + b) · A-2 = [SiNx + (√ 3) cosx] SiNx + 3 / 2-2 = sin? 2x + (√ 3) sinxcosx-1 / 2
=(1-cos2x)/2+(√3/2)sin2x-1/2=(√3/2)sin2x-(1/2)cos2x=sin2xcos(π/6)-cos2xsin(π/6)
=sin(2x-π/6)
Since f (a) = sin (2A - π / 6) = 1, 2A - π / 6 = π / 2, 2A = π / 2 + π / 6 = 2 π / 3,  a = π / 3
SΔABC=(1/2)bcsinA=(1/2)×2×4×sin(π/3)=2√3.

Given the vector → a = (2cosx / 2, Tan (x / 2 + π / 4)), → B = (√ 2Sin (x / 2 + π / 4), Tan (x / 2 - π / 4)), Let f (x) = → a · → B. whether there is a real number x ∈ [0, π], such that f (x) + F '(x) = 0, if it exists, then find the value of X; if not, prove it

Let f (x) + F '(x) = 0. F (x) = A. B = 2 √ 2 cosx / 2 sin (x / 2 + π / 4) + Tan (x / 2 + π / 4) Tan (x / 2 - π / 4) = 2 √ 2 cosx / 2 (√ 2 / 2 SiNx / 2 + √ 2 / 2cosx / 2) + (1 + TaNx / 2) / (1

Given that the opposite sides of the inner angles a, B, C in the acute triangle ABC are a, B, C, the vector m is defined=