If the function y = 2Sin square x (x + π / 4) - cos2x is known, then an equation of symmetry axis of the image of the function is A.x=-3x/8 B.x=-π/8 C.x=π/8 D.x=5π/8 Can you write the process 、、

If the function y = 2Sin square x (x + π / 4) - cos2x is known, then an equation of symmetry axis of the image of the function is A.x=-3x/8 B.x=-π/8 C.x=π/8 D.x=5π/8 Can you write the process 、、

Do you have an extra X in your question? If so, choose by = 2Sin? 2 (x + π / 4) - cos2x = 1-cos (2x + π / 2) - cos2x = 1 + sin2x-cos2x = 1 + √ 2Sin (2x - π / 4) the symmetry axis of trigonometric function, i.e. the line that makes the function get the maximum or minimum value. The period of the above function is t = 2

F (x) = SiNx ^ 4 + 2 times root sign 3 sinxcosx cosx ^ 4 find (1) function f (x) minimum positive period (2) function monotonicity in closed interval 0 to π closed interval Increasing interval

(1) (x) = SiNx ^ 4 + 2 + 2 + 3 SiNx cosx-cosx ^ 4 = SiNx ^ 4-cosx ^ 4 + 4 + (3 sin2x = √ 3sin2x + (SiNx ^ 2 + cosx ^ 2) (SiNx ^ 2-cosx ^ 2) (SiNx ^ 2-cosx ^ 2) (SiNx ^ 2-cosx ^ 2) (SiNx ^ 2-cosx ^ 2) = 2 [sin2x * (3 / 2) - cos2x * (1 / 2)] = 2 [sin2x * cos (π / 6) - cos2x * sin (π / 6)] = 2Sin (2x (2x - π π π - π - π / 6)] = 2Sin (2x (2x - π - π - π - π - 6)] = 2Sin (2x (2x/ / F (x) min

Let the maximum value of the function f (x) = cos2x + 2 √ 3sinxcosx (x belongs to R) be m and the minimum positive period be t;

By using asin α + bcos α = √ (a ^ 2 + B ^ 2) sin (α + φ) (Tan φ = B / a), t = Pi, M = 2

The minimum positive period of the function f (x) = sin4x + cos2x is___ .

y=sin4x+cos2x
=（ 1-cos2x
2）2+1+cos2x
Two
=cos22x+3
4=1+cos4x
Two
4+3
Four
=1
8cos4x+7
8．
∵ω=4，
The minimum positive period T = 2 π
4=π
2．
So the answer is: π
Two

If the function y = 2Sin ^ 2 (x + π / 4) - cos2x is known, then its period and a symmetric axis equation of the image are?

Simplify y = 2Sin ^ 2 (x + π / 4) - cos2x = 1-radical 2cos (2x + π / 4), so period T = π, symmetric axis equation, x = - π / 8 + n π, n is an integer

Given the function y = 2Sin ^ 2 (x + Pai / 4) - cos2x, then the equation of the symmetry axis of the period and the image is

Y = 2Sin ^ 2 (x + Pai / 4) - cos2x = 2 * (1-cos ^ (2x + π / 2)) / 2-cos2x = 1 + sin2x cosx = 1 + root 2Sin (x - π / 4). Do you understand it? I don't understand. Tell me which step it is. It seems that the trigonometric formula has been completely forgotten

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, π / 2), 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 ∈ [π / 2, π], calculate the value of x0

(1)T=2π/w=π w=2
When x = 0, y = 2cos (θ) = √ 3 cos θ = √ 3 / 2 θ = π / 6
y=2cos(2x+π/6)
(2) Is there a problem with the numbers in the following question? How can we say it very complicated

The known function f (x) = 2cosxsin (x + π / 3) + sinxcosx - √ 3sin? X, X ∈ R (1) Finding the minimum positive period of function f (x) (2) If there exists x ∈ [0,5 π / 12], the inequality f (x) > m is established, and the value range of real number m is obtained

f(x)=2cosxsin(x+π/3)-√3sin²x+sinxcosx
=2cosx(1/2*sinx+√3/2*cosx) -√3sin²x+sinxcosx
= sinxcosx+√3cos²x-√3sin²x+sinxcosx
=2 sinxcosx+√3(cos²x-sin²x)
=sin2x+√3cos2x
=2sin(2x+π/3)
(1)
Minimum positive period = 2 π / 2 = π
（2）
x∈[0,5π/12]
2x+π/3∈[π/3,7π/6]
sin(2x+π/3)∈[-1/2,1]
2sin(2x+π/3)∈[-1,2]
F (x) > m holds
M

[mathematics] the known function f (x) = 2cosxsin (x + π / 3) -√ 3sin ^ 2 (x) + sinxcosx Finding the minimum positive period of (1) function (2) Maximum and minimum values of functions (3) Increasing interval of function (4) The number of roots of the equation f (x) = x / (50 π)

f(x)=2cosx*[sinx*(1/2)+cosx*(√3/2)]-√3sin^2 x+sinxcosx
=sinxcosx+√3cos^2 x-√3sin^2 x+sinxcosx
=2sinxcox+√3(cos^2 x-sin^2 x)
=sin2x+√3cos2x
=2sin[2x+(π/3)]
So:
① Minimum positive period T = 2 π / 2 = π
② Max = 2, min = - 2
③ The increasing range is 2x + (π / 3) ∈ [2K π - (π / 2), 2K π + (π / 2)]
===> 2x∈[2kπ-(5π/6),2kπ+(π/6)]
===> x∈[kπ-(5π/12),kπ+(π/12)]（k∈Z）
④f(x)=x/(50π)
The number of roots is equivalent to the number of intersections of F (x) = 2Sin [2x + (π / 3)] and the line g (x) = x / (50 π)
The period of F (x) is π, and the number of intersections between F (x) and G (x) is 2 in each period
So, there are 100 roots

The known function f (x) = 2cosxsin (x + π / 3) -√ 3sin? X + sinxcosx (x belongs to R) The maximum value of 1 function and X when getting the maximum value Monotone decreasing interval of 2-function I'd like to ask you very much

Solution ∵ f (x) = 2cosx [sinxcos (π / 3) + cosxsin (π / 3)] - √ 3 (SiNx) ^ 2 + sinxcosx
=sinxcosx+√3(cosx)^2-√3(sinx)^2+sinxcosx
=sin2x+√3cos2x
=2[(1/2)sin2x+(√3/2)cos2x]
=2[sin2xcos(π/3)+cos2xsin(π/3)]
=2sin[2x+(π/3)]
When 2x + (π / 3) = π / 2, the function f (x) has a maximum value and f (x) = 2, where x = π / 12;
When 2x + (π / 3) = 3 π / 2, the function f (x) has a minimum value and f (x) = - 2. In this case, x = 7 π / 12;
The monotone decreasing interval of SiNx is 2K π + (π / 2) ≤ x ≤ 2K π + (3 π / 2);
∴2kπ+（π/2)≤2x+(π/3)≤2kπ+（3π/2)
The result is: K π + (π / 12) ≤ x ≤ K π + (7 π / 12)