The minimum positive period of the function f (x) = cos ^ 2 (2x - π / 6) is

The minimum positive period of the function f (x) = cos ^ 2 (2x - π / 6) is

f(x)=cos^2(2x-π/6)=[1+cos(4x-π/3)]/2 T=2π/4=π/2

The minimum positive period of the function f (x) = cos ^ 2x-1 / 2 is given

∵f(x)=cos^2x-1/2=(1/2)(1+cos2x)-1/2=(1/2)cos2x
The function f (x) = cos ^ 2x-1 / 2 minimum positive period T = 2 π / 2 = π

Let f (x) = 2sinxcosx cos (2x-6 π)? (1) find the minimum positive period of the function f (x)? (2) when x ∈ [0,2 π / 3], we can find the function f (x) = (2) when x ∈ [0,2 π / 3], we can find the function f (x)... (2) when x ∈ [0,2 π / 3] Let f (x) = 2sinxcosx cos (2x-6 π)? (1) find the minimum positive period of function f (x)? (2) when x ∈ [0,2 π / 3], find the maximum value of function f (x) and the value of X when the maximum value is obtained

f(x)=2sinxcosx-cos(2x-6π)
＝sin2x-cos2x
=√2sin(2x-π/2)
Minimum positive period = 2 π / 2 = π
When x ∈ [0,2 π / 3], 2x - π / 2 ∈ [- π / 2,5 π / 5]
Therefore, when 2x - π / 2 = π / 2, that is, x = π / 2
The maximum value of F (x) is equal to √ 2

Find the value range of the function y = cos2x + sinxcosx

y=cos2x+sinxcosx=1+cos2x
2+1
2sin2x=1
2(sin2x+cos2x)+1
Two
=
Two
2 (
Two
2sin2x+
Two
2cos2x)+1
2=
Two
2sin(2x+π
4)+1
Because of sin (2x + π)
4）∈[-1，1]
So the value range of the original function is [1]
2-
Two
2，1
2+
Two
2]

Find the value range of the function y = cos squared y = cos squared x + sinxcosx

First teach you to square it, or later you ask questions look very awkward. Hold down alt, then press the keypad 178
y=cos²x+sinxcosx
＝(1+cos2x)/2+sin2x/2
=(1/2)+(1/2)(sin2x+cos2x)
=(1 / 2) + (1 / 2) (radical 2) sin (2x + 45)
Because - 1 = so [(root 2) - 1] / 2 = < (1 / 2) + (1 / 2) (root 2) sin (2x + 45) < = [(root 2) + 1] / 2
The value range of the function is [[(radical 2) - 1] / 2, [(radical 2) + 1] / 2]

Find the value range of y = cos ^ 2x + cosxsinx That's the square of cosx

y=[2(cosx)^2+1+2cosxsinx-1]/2
=[2(cosx)^2-1+2cosxsinx+1]/2
=[cos(2x)+sin(2x)+1]/2
=[√2sin(2x+π/4)+1]/2
∵-1≤sin(2x+π/4)≤1
∴(1-√2)/2≤y≤(1+√2)/2

The value range of y = root 3sinx + cos

f（x）=√3sinx+cosx
=2[sinx*(√3/2)+cosx*(1/2)]
=2[sinx*cos(π/6)+cosxsin(π/6)]
=2sin(x+π/6)
So the range is [- 2,2]

The range of y = cos ^ 2x-4sinx is 【-4,4】?

The derivation of Y yields y '= - 2cosxsinx-4cosx = - 2cosx (SiNx + 2), only when x = KPI / 2, y' = 0
When x = KPI / 2, cosx = 0, - 4sinx = 4 or - 4 is the extreme value
Because y is continuous, the range is [- 4,4]

The range of y = 2-cos (2x - π / 6) x∈[π/6,3π/4]

The value range of y = 2-cos (2x - π / 6) x ∈ [π / 6,3 π / 4]
x∈[π/6,3π/4]
Then 2x - π / 6 ∈ [π / 6,4 π / 3]
From the cosine image, we can get the minimum value of COS (2x - π / 6) and the maximum value of y when 2x - π / 6 = π, that is, x = 7 π / 12
When x = π / 6 or x = 3 π / 4, the maximum value of COS (2x - π / 6) (root 3 / 2), that is, the minimum value of Y is 2-radical 3 / 2
Thus, the range of Y is obtained: y ∈ [2-radical 3 / 2,3]
It should be detailed enough,

Find the value range of the function y = cos ^ 2x SiNx on [0, π]

y=1-sin²x-sinx
=-(sinx+1/2)²+5/4
The opening is downward and the axis of symmetry SiNx = - 1 / 2
X belongs to [0, π]
Zero