The known function f (x) = Cos2 ω x+ The minimum positive period of 3sin ω xcos ω x (ω > 0) is π (I) find f (2) 3 π) (II) find the monotone increasing interval of function f (x) and the symmetric axis equation of its image

The known function f (x) = Cos2 ω x+ The minimum positive period of 3sin ω xcos ω x (ω > 0) is π (I) find f (2) 3 π) (II) find the monotone increasing interval of function f (x) and the symmetric axis equation of its image

(I) f (x) = 12 (1 + Cos2 ω x) + 32sin2 ω x = 12 + sin (2 ω x + π 6), because the minimum positive period of F (x) is π, 2 π 2 ω = π, and the solution ω = 1, so f (x) = sin (2x + π 6) + 12, so f (2 π 3) = - 12. (II) from 2K π - π 2 ≤ 2x + π 6 ≤ 2K π + π 2, (K ∈ z), K is obtained

Known function f (x)= The minimum positive period T = π of 3sin ω x · cos ω x-cos2 ω x, (ω > 0) 2． (I) find the value of real number ω; (II) if x is the minimum inner angle of △ ABC, find the value range of function f (x)

(I) because f (x)=
Three
2sin2ωx-1
2(1+cos2ωx)=sin(2ωx-π
6)-1
2，
So t = 2 π
2ω=π
2，∴，ω=2．
(II) because x is the minimum internal angle of △ ABC, X ∈ (0, π)
3]，
F (x) = sin (4x - π)
6)-1
So f (x) ∈ [- 1,1
2]．

It is known that the minimum positive period of the radical sign 3sinwxcoswx cos ^ 2wx + 3 / 2, (x ∈ R, w ∈ R) is π, and when x = π / 6, the function has a minimum value If the image of function y = 1-f (x) has an intersection point with the line y = a at [0, π / 2], find the range of real number a

f(x)=√3sinωxcosωx-(cosωx)^2+3/2
=(√3/2)sin2ωx-(cos2ωx+1)/2+3/2
=sin(2ωx-π/6)+1
Because t = π
So t = 2 π / 2 ω = π
Therefore, ω = 1
So f (x) = sin (2x - π / 6) + 1
Zero

F (x) = 4sin? 2wx-3 radical sign 3sinwxcoswx + cos? Wx is a periodic function with the least positive period based on the dichotomy 1. Find the symmetry axis equation of y = f (x) image 2. Find the monotone positive interval of y = f (x), the maximum value, the value of X when the maximum value

1) (x) = 4sin / 2wx-3 √ 3sinwcoswx + CoS / 2wx = 2 (1-cos2wx) - 3 √ 3 / 3 / 2 * 2 sinwxcoswx + 1 / 2 (1 + cos2wx) = 2-2cos2wx-3 √ 3 / 2sin2wx + 1 / 2 + 1 / 2 + 1 / 2cos2wx = - 3 √ 3 / 2sin2wx-3 / 2cos2wx + 5 / 2 = - 3sin (2wx + π / 6) + 5 / 2  is based on the π \\π (2wx + π / 6) + 5 / 2 5757 87 / 2 is the minimum positive period

The function f (x) = asin (x + π / 4) -√ 6cos (x + π / 3), when is f (x) an even function and when is an odd function Do not use the method of F (x) = f (- x), too boring

Open to get (√ 6 + √ 2a) / 2 * cosx + (√ 2a-3 √ 2) / 2 * SiNx
Elimination of sina even function by a = 3
Elimination of cosa odd functions by a = - 3

Is the function f (x) = root (x ^ 2-1) + root (1-x ^ 2) odd or even

It is both an odd function and an even function
To judge parity, we should first look at the definition domain
X ^ 2-1 > = 0 and 1-x ^ 2 > = 0
That can only be equal to 1-x ^ 2 = 0, so x = ± 1
So f (x) = 0 (x = ± 1)
F (- x) = - f (x) and f (- x) = f (x)
It is both an odd function and an even function

Is the function f (x) = [radical (1 + x ^ 2) + X-1] / [radical (1 + x ^ 2) + X + 1] odd or even What I don't understand is its definition. The answer in practice is that x belongs to all real numbers. I hope you can give me some advice,

The root sign (1 + x ^ 2) + X + 1 is not equal to 0, and the solved x is the definition field
Radical (1 + x ^ 2) + X + 1 > |x| + X + 1
X0
x> When = 0, the above formula is 2x + 1 > = 1
So x belongs to R
The first condition of judging odd function and even function is that the definition domain is symmetric

If f (x) is an even function, find the value of F (1 + radical 2) - f [(1-radical 2) 1]

Because f (x) is an even function
So (x) = F
So f [(1-root 2) 1] = f [- (1 + root 2)] = f (1 + root 2)
Therefore, the value of F (1 + radical 2) - f [(1-radical 2) 1] is 0

If f (x) = cos (2x + a) - radical 3sin (2x-a) is even function, find a

(x) = cos (- 2x + a) - √ 3sin (- 2x-a) = cos (2x-a) + and √ 3sin (2x + a) = f (x) = cos (2x + a) - √ 3sin (2x-a) cos (2x + a) - cos (2x-a) = √ 3 [sin (2x + a) + sin (2x-a)] - 2sin2xsina = √ 3 * 2sin2xcosasana / cosa = - √ 3 = tanaa = k ΠΠ 3 K, K is a = k ΠΠk, K is k ΠΠk, K is a = k ΠΠΠ Π Π integers

Let f (x) = radical 3sin (x + a) + cos (x-a) be even function, a belongs to (0, PAI), find the value of A

If f (x) is an even function, then for any x there is f (x) = f (- x),
Therefore, the radical 3sin (x + a) + cos (x-a) = 3sin (- x + a) + cos (- x-a)
The root sign 3sin (x + a) + cos (x-a) = - root 3sin (x-a) + cos (x + a)
Radical 3sin (x + a) + Radix 3sin (x-a) = cos (x + a) - cos (x-a)
Two angle sum difference formula expansion and simplification
2 root sign 3sinx * cosa = 2sinx * Sina
The result shows that COTA = root 3 / 3, and a belongs to (0, PAI),
So a = Pie / 3