Given the function f (χ) = sin (2x + π / 6) + sin (2x - π / 6) + cos2x + 1 (x ∈ R), The minimum (2x) symmetric (x, x) + s (2 x, x) + s (2 x, x) + s (2 x, x) + s (2 x, x) + s (2 x, x) are given

Given the function f (χ) = sin (2x + π / 6) + sin (2x - π / 6) + cos2x + 1 (x ∈ R), The minimum (2x) symmetric (x, x) + s (2 x, x) + s (2 x, x) + s (2 x, x) + s (2 x, x) + s (2 x, x) are given

If (x) = (√ 3 / 2) sin2x + (1 / 2) cos2x + (1 / 2) cos2x + (√ 3 / 2) sin2x - (1 / 2) cos2x + cos2x + 1 = √ 3sin2x + cos2x + 1 = 2Sin (2x + π / 6) + 1 cycle t = 2 π / 2 = π symmetry axis: 2x + π / 6 = π / 2 + K π, the symmetry axis is: x = - π / 6 + K π / 2, K ∈ Z, symmetry center: 2x + π / 6 = k π, K ∈ Z symmetry center: 2x + π / 6 = k π, 6 = k π, K ∈ Z symmetry center: 2x + π / 6 = k π, the center of symmetry is: (...)

The known function f (x) = 2cos2x + cos (2x + π) 2），x∈R (1) Find the minimum positive period of F (x); (2) Find the monotone increasing interval of F (x); (3) If f (α) = 3 4. Find the value of SiN4 α

(1) If f (x) = 2cos2x + cos (2x + π 2) = 1 + cos2x-sin2x = 1 + 2cos (2x + π 4), the minimum positive period of F (x) t = 2 π 2 = π. (2) from (2x + π 4) ∈ [- π + 2K π, 2K π], X ∈ [K π − 5 π 8, K π − π 8] (K ∈ z) &

The function f (x) = 2cos? - cos (2x + π / 2) is known. The minimum positive period and monotone decreasing interval of function f (x) are obtained

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Let f (x) = 2cos squared cos (2x PAI-2), find the value of F (pi-8), find the minimum positive period and monotone decreasing interval of function f (x)

f(x)=2cos²x -cos(2x+π/2)=1+cos(2x) +sin(2x)=1+√2 sin(2x+π/4)
f(π/8)=1+√2sin(π/2)=1+√2 Tmin=2π/2=π
2K π + π / 2 ≤ 2x + π / 4 ≤ 2K π + 3 π / 2K π + π / 8 ≤ x ≤ K π + 5 π / 8 [K π + π / 8, K π + 5 π / 8]

Function f (x) = cos ^ 2x + 2cos ^ 2 x / 2 monotonically increasing interval

f(X)=cos^2x+2cos^2 x/2=f(X)=cos^2x+cos x+1=(cosx+1)^2
Cot = 1

Given the function f (x) = √ 3 / 2sin2x cos ^ 2x-1 / 2, X ∈ R, find the function f (the minimum positive period and monotone increasing interval of x) online and so on! Given the function f (x) = √ 3 / 2sin2x cos ^ 2x-1 / 2, X ∈ R, find the function f (the minimum positive period and monotone increasing interval of x)

f（x）＝√3/2sin2x-cos^2x-1/2
=√3/2sin2x-1/2*(1+cos(2x)) -1/2
=√3/2sin2x-1/2* cos(2x)-1
= sin(2x-π/6)-1
The minimum positive period of a function is 2 π / 2 = π
2kπ-π/2≤2x-π/6≤2kπ+π/2,k∈Z.
kπ-π/6≤x≤kπ+π/3,k∈Z.
So the monotone increasing interval of the function is [K π - π / 6, K π + π / 3], K ∈ Z

Find the minimum value and the minimum positive period of the function f (x) = √ 3 / 2sin2x cos ^ 2x-1 / 2 (x ∈ R)

cos2x=2cosx^2-1
cosx^2=(cos2x+1)/2
f(x)=√3／2sin2x-cos^2x-1／2 =√3／2sin2x-1/2cos^2x-1=sin(2x-pai/6)-1
So the minimum value = - 1-1 = - 2
The minimum period of positive π = 2 π

Given the function f (x) = cos (π / 3 + x) * cos (π / 3-x), G (x) = 1 / 2sin2x-1 / 4, find the minimum positive period of F (x)

F (x) = cos (π / 3 + x) * cos (π / 3-x) = (1 / 2cosx radical 3 / 2sinx) (1 / 2cosx + radical 3 / 2sinx)
=1/4(cos²x-3sin²x)
=1/4(2cos2x-1)
T=2π/2=π

If the function f (x) = - cos2x + 1 2 (x ∈ R), then f (x) is () A. The minimum positive period is π Odd function of 2 B. Odd functions with minimum positive period π C. The minimum positive period is φ = π Even function 3 D. Even functions with minimum positive period π

Function f (x) = - cos2x + 1
2=-1+cos2x
2+1
2=-1
2cos2x，
∵ω=2，∴T=2π
2=π，
Cos2x is even function,
Then the function is an even function whose minimum positive period is π
Therefore, D is selected

Let f (x) = cos (2x - π / 3) - cos2x-1 How to simplify it?

f(x)=cos(2x-π/3）-cos2x-1
=cos2x*cos(π/3）+ sin2x*sin(π/3） - 2cos2x*cos(π/3）-1
= - [cos2x*cos(π/3）- sin2x*sin(π/3）] -1
=-cos(2x+π/3）-1