The function f (x) = sinxcosx + root 3cos ^ 2x - radical 3 / 2 Maximum increment interval of minimum positive period

The function f (x) = sinxcosx + root 3cos ^ 2x - radical 3 / 2 Maximum increment interval of minimum positive period

f(x)=sinxcosx+√3(cosx)^2-√3/2
=(1/2)sin2x+(√3/2)cos2x
=sin2xcosπ/3+cos2xsinπ/3
=sin(2x+π/3)

Let f (x) = (cosx) ^ 2 + root sign 3 * sinxcosx + a if x ∈ [0, π / 2], the minimum value of F (x) is 2, and find the value of A

f(x)=(cos2x+1)/2+√3*sin2x/2+a
=sin(2x+π/6)+1/2+a
So when x ∈ [0, π / 2]
that
2x+π/6∈[π/6,7π/6]
So the minimum
f(x)=sin(7π/6)+1/2+a
=-1/2+1/2+a
=a=2
So, a = 2

Given the function y = (cosx) ^ 2 + (radical 3) sinxcosx + 1, X ∈ R Given the function y = (cosx) ^ 2 + (radical 3) sinxcosx + 1, X ∈ R, How can the image of this function be obtained from the image of y = sin2x (x ∈ R)?

According to the title,
Given the function y = cos 2 x + (√ 3) sinxcosx + 1, X ∈ R,
So, simplify
y=cos²x+(√3)sinxcosx+1
=(cos(2x)+1)/2+(√3/2)sin(2x)+1
=(1/2)cos(2x)+(√3/2)sin(2x)+3/2
=sin(2x+π/6)+3/2
=sin(2(x+π/12))+3/2
So,
Images with y = sin2x (x ∈ R)
First shift π / 12 units in the negative direction of the x-axis
Move 3 / 2 units in the positive direction of the y-axis
The image of y = sin (2 (x + π / 12)) + 3 / 2 is obtained

Let the function FX = 6cos square X-2 radical sign 3sinxcosx. Find the minimum positive period sum range of FX

2-2 √ 3sinxcos x = 6 * (cos2x + 1) / 2 - √ 3sin2x = 3cos 2x = 3cos 2x - √ 3sin2x + 3 = 2 √ 3 (3 / 3 / 2cos2x-1 / 2sin22x) + 3 = 3 = 2 √ 3sin (π / 3-2x) + 3 = 3 = - 2 √ 3sin (2X - π / 3 / 3) + 3, so the maximum value of F (x) is 2 √ 3 + 3, and the minimum positive period is π. If the acute angle α is α, if the acute angle is α, then the minimum positive period is π. If the acute angle is α, if the acute angle is α, if the acute angle α is α, the minimum positive period is π. If the acute if f (α) = 3 - 2 √

The value range of the function y = cos ^ 2 + root sign 3sinxcosx on the closed interval between - π / 6 and π / 4 is

Y = cos ^ 2 x + radical 3sinxcosx
=1/2+1/2 cos2x+√3/2 sin2x
=1/2+sin(2x+π/6)
-π/3≤2x≤π/2
-π/6≤2x+π/6≤2π/3
-1/2≤sin(2x+π/6)≤1
0≤y≤3/2

Is the value range of the function y = radical 3sinxcosx + cos ^ 2x-1 / 2 in [0, π / 2]? A.[-1,1] B.[1/2,1] C.[0,1] D.[-1/2,1] Radical 3

Y = √ 3sinxcosx + cos ^ 2x-1 / 2 = (√ 3 / 2) sin2x + (1 / 2) (cos2x + 1) - 1 / 2 = (√ 3 / 2) sin2x + (1 / 2) cos2x = sin (2x + π / 6) because x belongs to [0, π / 2], 2x + π / 6 belongs to [π / 6,7 π / 6], so the minimum value is sin (7 π / 6) = - 0

Find the period and range of the function f (x) = 3sinxcosx-3 root sign 3cos ^ 2x + 3 radical sign 3

The period and range of the function f (x) = 3sinxcosx - (3 √ 3) (cosx) ^ 2 + 3 √ 3 is the period and value domain of the function f (x) = 3sinxcosx - (3 √ 3) (cosx) ^ 2 + 3 √ 3 = 3 [(sin2x) / 2] - (3 √ 3) [(cos2x + 1) / 2] + 3 √ 3 = {3 (1 / 2) sin2x - (√ 3 / 2) cos2x} + (3 √ 3) / 2 = 3 = 3sin (2x-60-60-60-60-3-3-3 / 3) / 2 = 3sin (2x-60-60-60-60-60-3-3-3-3 / 2 = 3sin (2x-60-60-?) + (3 √ 3) /

Let f (x) = 2cosxsin (x + Π / 3) - [(radical 3) / 2] Find the minimum positive period of function f (x) Would you please write down the calculation process and result?

f(x)=2cosx(sinxcosπ/3+cosxsinπ/3)-√3/2
=sinxcosx+√3cos²x-√3/2
=1/2*sin2x+√3(1+cos2x)/2-√3/2
=1/2*sin2x+√3/2*cos2x
=sin2xcosπ/3+cos2xsinπ/3
=sin(2x+π/3)
T=2π/2=π

The known function f (x) = 2cosxsin (x + π / 3) -√ 3 (SiNx) ^ 2 + sinxcosx Find the increasing interval of (1) function f (x) (2) The maximum and minimum of function f (x) (3) The number of roots of the equation f (x) = x / 50 π

Note: all irrational numbers "Pai" are represented by PI: F (x) = 2cosxsin (x + π / 3) -√ 3 (SiNx) ^ 2 + sinxcosx = 2cosx (SiNx / 2 + √ 3cosx / 2) -√ 3 (1-cos2x) / 2 + sin2x / 2 = sinxcosx + √ 3 (1 + cosx) / 2 - √ 3 / 2 + √ 3cos2x / 2 + sin2x / 2 = sin2x + √ 3cos2x = 2Sin (2x + pi / 3) (1)

Hurry up Given that f (x) = sin ^ 2x + radical 3 * sinxcosx + 2cos ^ 2x, find (1) f (x) the minimum positive period (2), (3), (4), (5) (2) The maximum value of F (x) monotonically increasing interval (3) f (x) and the corresponding x value (4) f (x) symmetry axis equation and symmetry center (5) can be obtained by how to transform the image with y = sin2x

(x) = sin ^ 2x + root number 3 * sinxcos x + 2cos ^ 2x = 1 + √ 3 / 2sin22x + cos 2x = 1 + 1 + √ 3 / 2sin22x + (1 / 2) (1 + cos2x) = sin (2x + π / 6) + 3 / 2 (1) t = π (2) [K π - π / 3, K π + π + π / 6] (3) maximum value is 5 / 2, at this time x = k π + π / 6 (4) f (x) symmetric axis equation is x = k π + π + π + π + 6 (4) f (x) symmetric axis equation is x = k π + π + π + π + π + π + π + π + π + π + π/ 6 or x = k π - π