The minimum positive period of the function y = 2cos ^ 2x sinxcosx is

The minimum positive period of the function y = 2cos ^ 2x sinxcosx is

The solution is y = 2cos ^ 2x - sinxcosx
=2cos^2x-1-1/2*2sinxcosx+1
=cos2x-1/2sin2x+1
=√5/2(2/√5cos2x-1/√5sin2x)+1
=√5/2cos（2x+θ）+1
The minimum positive period is t = 2 π / 2 = π

The minimum positive period of the function y = 2cos ^ 2x 1 is seek

y＝2cos^2x+ 1
=1+cos2x+1
=2+cos2x
therefore
Period = 2 π / w = 2 π / 2 = π

The minimum positive period of the function y = 2cos2x + 1 (x ∈ R) is______ .

y=2cos2x+1=1+cos2x+1=cos2x+2，
∵ω=2，∴T=2π
2=π．
So the answer is: π

Find the period of the following function: y = - 2cos (- 1 / 2x-1) Find the period of the following functions: ① y=-2cos(-1/2x-1) ② y=│sin2x│ ③ y=cos3x+sin2x

For example, the period of a function like y = asin (BX + C) or y = ACOS (BX + C) is | 2 π / b | with absolute value, that is, if the image below the X axis is turned up, the period must be divided by 2. Therefore, the period of ① is | 2 π / (- 1 / 2) | 2 π / 2 = π / 2, and ③, the period of cos3x is | 2 π / 3 = 2 π / 3sin2x

Given the function f (x) = √ 3sin (2x - π / 6) + 2cos 2 (x - π / 12), find the minimum positive period and maximum value of function f (x)

Cos 2 (x - π / 12) is reduced to (2cos (x - π / 6) + 1) / 2 by the formula of double angle, and then combined with √ 3sin (2x - π / 6)

Given the function f (x) = 1 - √ 3sin (2x) + 2cos 2 (x), find the value range of F (x)

f（x）=1-√3sin（2x）+2cos²（x）
＝cos2x－√3sin2x＋2
＝2cos(2x+π/3）＋2
∵2cos(2x+π/3）∈［－2,2］
Therefore, the value range of F (x) is [0,4]

Let FX = cos (2x - 4 π / 3) + 2cos? X ① Find the maximum value of FX, and write the set of values of X when FX takes the maximum value ② We know that the opposite sides of angles a, B and C in the triangle ABC are ABC respectively. If f (B + C) = 3 / 2, B + C = 2, find the minimum value of A

(1)
f(x)=cos﹙2x－4π／3﹚＋2cos²x
=cos2xcos4π/3+sin2xsin4π/3+1+cos2x
=1/2cos2x-√3/2sin2x+1
=cos(2x+π/3)+1
When 2x + π / 3 = 2K π, K ∈ Z
When x = k π - π / 6, K ∈ Z
The maximum value of F (x) is 1 + 1 = 2
The set of X is {x | x = k π - π / 6, K ∈ Z}
II.
If f (B + C) = 3 / 2
That is cos [2 (B + C) + π / 3] + 1 = cos [(2 π - 2A) + π / 3] + 1
=cos(2A-π/3)+1=3/2
∴cos(2A-π/3)=1/2
∵ 0

Find the minimum positive period, maximum value and sum of F (x) = cos (2x + Pai / 3) + 2Sin ^ 2x Monotone decreasing interval. Thank you very much~

F (x) = (1 / 2) cos2x - (√ 3 / 2) sin2x + 1-cos2x = - [(√ 3 / 2) sin2x + (1 / 2) cos2x] + 1 = - [sin (π / 3) sin2x + cos (π / 3) cos2x] + 1 = - cos (2x - π / 3) + 1. Therefore, t = 2 π / 2 = π Max is 2 monotonic decreasing interval: π + 2K π

Function y = 1 The minimum positive period of 2sin22x is () A. π Two B. π C. 2π D. 4π

∵ y = 1
2sin22x=1
2•1−cos4x
2=1−cos4x
4，
The period T = 2 π
4=π
2．
Therefore, a

The minimum positive period of the function y = 1 / 2Sin ^ 2x

cos4x=1-2sin²2x
So y = 1 / 2Sin? 2x = 1 / 2 * (1-cos4x) / 2
=-1/4*cos4x+1/4
So t = 2 π / 4 = π / 2