Let f (x) = cosx ^ 2-sinx ^ 2 + (2 radical 3) sinxcosx + 1 1. Find the minimum positive period and minimum value of F (x) 2. If f (a) = 2 and a belongs to [π / 4, π / 2], find the value of A

Let f (x) = cosx ^ 2-sinx ^ 2 + (2 radical 3) sinxcosx + 1 1. Find the minimum positive period and minimum value of F (x) 2. If f (a) = 2 and a belongs to [π / 4, π / 2], find the value of A

F (x) = cosx ^ 2-sinx ^ 2 + (2 radical 3) sinxcosx + 1
=Cos2x + Radix 3sin2x + 1
=2 (radical 3 / 2 sin2x + 1 / 2 cos2x) + 1
= 2( sin2xcosπ/6 + cos2xsinπ/6 ) + 1
= 2 sin(2x+π/6) + 1
Minimum positive period = 2 π / 2 = π
Minimum = 2 * (- 1) + 1 = - 1
f（a）=2
2 sin(2a+π/6) + 1 = 2
sin(2a+π/6) = 1/2
A belongs to [π / 4, π / 2]
2a∈【π/2,π】
2a+π/6∈【2π/3,7π/6】
2a+π/6 = 5π/6
a = π/3

Find the monotone decreasing interval of the function f (x) = cosx - (radical 3) SiNx in [0,2 π]

solution
f(x)=cosx-√3sinx
=2[(1/2)cosx-(√3/2)sinx]
=2cos(x+π/3)
Let 2K π ≤ x + π / 3 ≤ 2K π + π to obtain 2K π - π / 3 ≤ x ≤ 2K π + 2 π / 3
∵ x ∈ [0,2 π], K can obtain 0 ≤ x ≤ 2 π / 3, K take 1 to get 5 π / 3 ≤ x ≤ 2 π
The monotone decreasing interval of F (x) is [0,2 π / 3] ∪ [5 π / 3,2 π]

The minimum positive period of the function f (x) = cos2x is______ .

f（x）=cos2x，
∵ω=2，∴T=2π
2=π．
So the answer is: π

Find the minimum positive period of the function f (x) = √ (1-cos2x) + √ (1 + cos2x)

1-cos2x
=1-(1-2sin²x)
=2sin²x
1+cos2x
=1+2cos²x-1
=2cos²x
So f (x) = √ 2 (| SiNx | + | cosx |)
f(x+π/2)=√2(|cosx|+|-sinx|)=f(x)
T=π/2

The minimum positive period of the function f (x) = cos2x is______ .

f（x）=cos2x，
∵ω=2，∴T=2π
2=π．
So the answer is: π

As shown in the figure, the image of the function y = 2cos (ω x + θ) (x ∈ R, 0 ≤ θ ≤) intersects the point (0, radical 3) with the minimum positive period of π (1) Find the values of θ and ω (2) Given the point a (2,2 / π), the point P is the point on the image of the function, and the point Q (x0, Y0) is the midpoint of PA. when Y0 = 2 / radical 3, x0 ∈ [2 / π, π], calculate the value of x0 2/π=2÷π

2π/ω=π ω=2
Y = 2cos (2x + θ) x = 0, y = radical 3
Cos θ = radical 3 / 2 θ = π / 6

The image of the function f (x) = 2cos (ω x + θ) (x ∈ R, 0 ≤ θ ≤ π - 2, ω > 0) intersects the point (0, radical 3) with the minimum positive period of π (1) The image of the function f (x) = 2cos (ω x + θ) (x ∈ R, 0 ≤ θ ≤ π - 2, ω > 0) intersects the point (0, radical 3) with the minimum positive period of π (1) To find the value of θ and ω; (2) to find the monotone increasing interval of function f (x)

1 T = 2 π / ω = π = = > ω = 2  image and y-axis intersection point (0, √ 3), and  2cos θ = √ 3, = = > cos θ = = > cos θ = (3 / 2 = = > θ = 2K π ± π / 6, K ∈ Z

Given the function f (x) = 2sinx (SiNx + cosx) - 1, the function f (x) is drawn in the plane rectangular coordinate system in [0180]

f(x)=2sinx(sinx+cosx)-1
=2sinx*sinx+2sinx*cosx-(sinx*sinx-cosx*cosx)
=sinx*sinx-cosx*cosx+2sinx*cosx
=-cos2x+sin2x
=sin2x-cos2x
=√2((sin2x*(1/√2)-cos2x*(1/√2)))
=√2((sin2x*cos(π/4)-cos2x*sin(π/4)))
=√2((sin(2x-π/4))

Definition domain of y = LG (2sinx-1) + radical - 2cosx

The requirement of this question is
2sinx-1＞0,
-2cosx＞0,
In other words, SiNx > 0.5, cosx < 0
It is known that x is the intersection of (2k π + π / 6,2k π + 5 / 6 π) and (2k + 1 / 2 π, 2K + 3 / 2 π)
That is (2k + 1 / 2 π, 2K π + 5 / 6 π)

Given the function f (x) = - radical (2) sin (2x + π / 4) + 6sinxcosx-2cosx ^ 2 + 1, X belongs to R, find the minimum positive period of F (x)

f(x)=-√2sin(2x+π/4)+6sinxcosx-2cosx^2+1,
=-sin2x-cos2x+3sin2x-cos2x,
=2sin2x-2cos2x
=2√2sin(2x-π/4)
Minimum positive period T = π