Let f (x) = √ 3cos? X + sinxcosx - √ 3 / 2 (1) find the minimum positive period T of function f (x), and find the monotone increasing interval of function f (x); (2) Find the sum of all x in [0,3 π) such that f (x) reaches the maximum value

Let f (x) = √ 3cos? X + sinxcosx - √ 3 / 2 (1) find the minimum positive period T of function f (x), and find the monotone increasing interval of function f (x); (2) Find the sum of all x in [0,3 π) such that f (x) reaches the maximum value

(1) (x) = (x) = (3 / 2) [(COS (2x) + 1] + (1 / 2) SiNx (2x) + 1] + (1 / 2) sin (2x) - √ 3 / 2 = (3 / 2) cos (2x) + (1 / 2) sin (2x) + (1 / 2) sin (2x) = cos (π / 6) cos (2x) + sin (π / 6) sin (2x) = cos (2x - π / 6) minimum positive cycle T = 2 π / 2 = π 2K π π - π - π - π - π - π - π - π, π, π, π, π, π, π, π, π, π, π, π, π, π, when - π / 6 ≤ 2K π, f (x) is simple

Let f (x) = sin2x-2cos 2 X. find the minimum positive period of F (x) Finding the minimum value of F (x) and the set of X when removing the minimum value

f(x)=sin2x-2sin^2x
=sin2x-(1-cos2x)
=sin2x+cos2x-1
=√ 2 (√ 2 / 2 * sin2x + √ 2 / 2cos2x) - 1 (extract √ 2)
=√ 2 (sin2xcos π / 4 + cos2xsin π / 4) - 1
=√2sin(2x+π/4)-1
(1) The minimum positive period of function f (x) t = 2 π / 2 = π
(2) When 2x + π / 4 = 2K π + π / 2, K ∈ Z, that is, x = k π + π / 8
The set of X is {x | x = k π + π / 8, K ∈ Z}

Let f (x) = 2cos? X + sin2x + A + 1 (a ∈ R). (1) find the minimum positive period sum of function f (x) Let f (x) = 2cos? X + sin2x + A + 1 (a ∈ R) (1) Find the minimum positive period and monotone increasing interval of function f (x); (2) When x ∈ [0, Π / 3], the maximum value of F (x) is 3. Find the value of A

1)f(x)=(1+cos2x)+sin2x+a+1=sin2x+cos2x+a+2=√2sin(2x+π/4)+a+2
Minimum positive period T = 2 π / 2 = π
Monotone increasing interval: 2K π - π / 2=

An urgent known function f (x) = sin2x-2cos 2 x (x ∈ R) (1) find the minimum positive period of function f (x) (2) When x ∈ (0, tt / 2), find the maximum value of function f (x) and the corresponding value of X

（1）f(x)=sin2x-2cos²x=sin2x-（1+cos2x）=sin2x-cos2x-1=√2sin（2x-π/4）-1
T=2π/2=π
(2) X belongs to (0, π / 2) and 2x - π / 4 belongs to (- π / 4,3 π / 4)
When 2x - π / 4 = π / 2, that is, x = 3 π / 8, the maximum value of F (x) is √ 2-1

Find the minimum positive period of the function f (x) = 2cos 2 (x + π of 12) + sin2x,

=Cos (2x + π / 6) + sin2x = cos2xcos π / 6-sin2xsin π / 6 + sin2x = (√ 3 / 2) cos2x - (1 / 2) sin2x + sin2x = (√ 3 / 2) cos2x + (1 / 2) sin2x = cos2xcos π / 6 + sin2xsin π / 6 = cos (2x - π / 6), so the minimum positive period of F (x) is π

The known function f (x) = sin2x-2sin2x (I) find the minimum positive period of function f (x); The minimum value of (x) and (x) is obtained

(I) because f (x) = sin2x - (1-cos2x)=
2sin（2x+π
4）-1，
So the minimum positive period of the function f (x) is t = 2 π
2=π
(II) from (I), when 2x + π
4=2kπ−π
2，
That is, x = k π − π
When 8 (K ∈ z), the minimum value of F (x) is −
2−1；
Therefore, when the function f (x) takes the minimum value, the set of X is: {x | x = k π - π
8，k∈Z}

Given the function FX = 2Sin? X + sin2x-1, find the maximum value of the function

The function FX = 2Sin? X + sin2x-1
=sin2x-cos2x
=√2sin(2x-π/4)
Maximum = √ 2

Let f (x) = (sin2x + cos2x) 2-2sin22x (I) find the minimum positive period of F (x); (II) if the image of function y = g (x) is shifted from the image of y = f (x) to the right π If x ∈ [0, π 4] Find the maximum and minimum values of y = g (x)

（Ⅰ）∵f（x）=（sin2x+cos2x）2-2sin22x
=sin22x+2sin2xcos2x+cos22x-（1-cos4x）
=1+sin4x-1+cos4x=sin4x+cos4x=
2sin（4x+π
4），
The minimum positive period of the function f (x) is 2 π
4=π
2；
(II) according to the meaning of the title, y = g (x)=
2sin[4（x-π
8）+π
4]=
2sin（4x-π
4），
∵0≤x≤π
4，∴-π
4≤4x-π
4≤3π
4，
When 4x - π
4=π
2, that is, x = 3 π
At 16, G (x) is the maximum
2；
When 4x - π
4=-π
When x = 0, the minimum value of G (x) is - 1

The known function f (x) = sin? X + 2sinxcosx + 3cos? X, X ∈ R Given the function f (x) = sin? 2x + 2sinxcosx + 3cos? X, X ∈ RR, we obtain the following results: (1) The minimum positive period and maximum value of function f (x) (2) The monotone increasing interval of function f (x)

f(x)=1+2cos^2x+sin2x
=(cos2x+2)+sin2x
=(cos2x+sin2x）+2
=√2sin(2x+π/4）+2
(1) The minimum positive period of the function f (x) is π and the maximum value is √ 2 + 2;
(2)2kπ-π/2

Let f (x) = sin 2 x + 2 SiN x cos x + 3 cos 2 x, X ∈ R, find the monotone increasing interval of function f (x) on the interval [0, π / 2]

f(x)=1+sin2x+2cos^2x
=1+sin2x+1+cos2x
=√2sin(2x+π/4)+2
2kπ-π/2≤2x+π/4≤2kπ+π/2
kπ-3π/8≤x≤kπ+π/8
[0,π/8]