Let the equation SiNx + radical 3cosx = a have two distinct real roots X1 and X2 in the interval (0,2). Find the value range of a and the value of X1 + x2

Let the equation SiNx + radical 3cosx = a have two distinct real roots X1 and X2 in the interval (0,2). Find the value range of a and the value of X1 + x2

SiNx + √ 3cosx = a SiNx * 1 / 2 + √ 3cosx / 2 = A / 2 sin (x + π / 3) = A / 2 when - 2

A = (radical 3cosx, cosx-1) B = (SiNx, cosx + 1) function f (x) = AB + 1 / 2 find f (x) period

f(x)=√3cosxsinx+(cosx-1)(cosx+1)+1/2
=√3cosxsinx+cos²x-1+1/2
=√3/2sin2x+(1+cos2x)/2-1/2
=√3/2sin2x+1/2cos2x
=sin(2x+π/6)
T=2π/2=π
Period π

Given the vector a = (SiNx, root 3cosx), vector b = (cosx, cosx), find the function f (x) = vector a · vector B, (1) Find the period and increasing interval of F (x) (2) if vector a ⊥ vector B, find the solution set of X

f（x）=a.b=sinxcosx+√3cosx^2=1/2sin2x+√3(cos2x+1)/2=sin(2x+π/3)+√3/2
(1) Increasing range: - π / 2 + 2K π

Let f (x) = SiNx (radical 3cosx SiNx), find the minimum positive period of function f (x)

The original formula can be reduced to: √ 3sinxcosx sin? X = (√ 3 / 2) sin2x + (cos2x) / 2-1 / 2 = sin (2x + π / 6) - 1 / 2
So the minimum positive period is 2 π / 2 = π

Given the function f (x) = SiNx radical 3cosx, find the minimum positive period of function f (x)

F (x) = SiNx radical 3cosx
=2 (1 / 2sinx radical 3 / 2cosx)
=2(sinxcosπ/3-cosx sinπ/3)
=2sin(x-π/3)
Minimum positive period: 2 π

Given the function f (x) = SiNx ^ 2 + 2 radical sign 3sinxcosx + 3cosx ^ 2, find the minimum proof period of function f (x)

f(x)=sin²x+2√3sinxcosx+3cos²x=1+2√3sinxcosx+2cos²x
=1+√3sin2x+1+cos2x
=2+√3sin2x+cos2x
=2+2sin(2x+π/6)
The period of the function is 2 π / 2 = π

Given the vector a (SiNx, - 1) vector b (radical 3cosx, - 1 / 2) function f (x) = (a + b) A-2, find the minimum positive period

A (SiNx, - 1), B (√ 3cosx, - 1 / 2) can be obtained by dissolving
a+b=(sinx+√3cosx,-3/2)
(a+b)a=sin^2x+√3sinxcosx-1/2
According to the trigonometric function formula
f(x)=1/√7/4sin(2x+φ)
Period T = 2 π / ω = 2 π / 2 = π
If you don't understand, ask again,

The monotone increasing interval of the function f (x) = SiNx minus the root sign 3cosx (x belongs to [negative distribution, 0]) is urgent

f(x)
＝sinx－√3cosx
＝2(sinx•1/2－cosx•√3/2)
＝2sin(x－π/3)
It consists of: - π / 2 + 2K π ≤ X - π / 3 ≤ π / 2 + 2K π
The result is: π / 6 + 2K π ≤ x ≤ 5 π / 6 + 2K π
∵x∈[－π,0]
Take the intersection and get: [- π / 6,0]

Function y = SiNx- The monotone increasing interval of 3cosx is______ .

∵y=sinx-
3cosx=2sin（x-π
3）
If 2K π - π
2≤x-π
3≤2kπ+π
2，k∈Z
Then 2K π - π
6≤x≤2kπ+5π
6，(k∈Z)
So the function y = SiNx-
The monotone increasing range of 3cosx is [2K π - π
6，2kπ+5π
6](k∈Z)
So the answer is: [2K π − π
6，2kπ+5π
6](k∈Z)

Let f (x) = SiNx radical 3cosx (x belongs to [- π, 0]) monotonically increasing interval is

f(x)=2×(1/2sinx-√3/2cosx)
=2×(cosπ/3sinx-sinπ/3cosx)
=2sin(x-π/3)
x∈【-π,0】
x-π/3∈【-4π/3,-π/3】
It increases when X - π / 3 ∈ [- π / 2, - π / 3]
Namely
x∈【-π/6,0】
That is, the increasing interval is [- π / 6,0]