Given the function f (x) = sin (2x + π / 6) + sin (2x + π / 6) + cos2x, find the minimum positive period of function y = f (x) Speed! Explain! Process!

f(x)=sin(2x+π/6)+sin(2x+π/6)+cos2x
=2sin(2x+π/6)+cos2x
=2[sin2xcosπ/6+cos2xsinπ/6]+cos2x
=√3sin2x+cos2x+cos2x
=√3sin2x+2cos2x
The minimum positive period of the function y = f (x)
Minimum positive period: π

The function f (x) = sin (2x + π / 6) + sin (2x - π / 6) - cos2x + a (a is a real number and belongs to R) By simplifying, f (x) = 2Sin (2x - π / 6) + A is obtained If x belongs to [π / 4, π / 2], the minimum value of F (x) is - 2

You're going to be simplified,
The minimum value is obtained at π / 4, where 2x - π / 6 = 5 π / 6, f (π / 4) = 1 + a = - 2, a = - 3

Let f (x) = sin (2x + φ), where φ is a real number, if f (x) ≤| f (π / 6) | holds for X ∈ R, and f (π 2) > F Why is the fifth line "f (π / 2) > F (π), i.e. sin φ < 0"?

The known function f (x) = sin (2x + φ), the known function f (x) = sin (2x + φ), where φ is a real number, if f (x) ≤| f (f (π / 6) | | for X ∈ r holds, and f (π / 2) > F (π (π / 2) > F (π), then the monotone increasing interval of F (x) is analytic: ∵ function f (x) = sin (2x + φ), f (x) function f (x) = sin (2x + φ), f (x) ≤ | f (π / 6) | for X ∈ R | f (x) x (x) take the maximum value of x = π / 6) | f (f (f (π / it's a good idea

Given the function f (x) = sin (2x + π / 6) + sin (2x - π / 6) - cos2x + A, find the monotone decreasing interval of the minimum positive periodic function FX of the function y = f (x), and the picture takes precedence

f(x)=2sin2xcos(π/6)-cos2x+a
=2sin2xcos(π/6)-2cos2xsin(π/6)+a
=2sin(2x-π/6)+a
Minimum positive period T = 2 π / 2 = π
The monotone decreasing interval is 2K π + π / 2=

We know that (2x s) = 2x F 2) (I) find f (π) 8) Value of; (II) find the minimum positive period and monotone decreasing interval of function f (x)

(Full Score: 13 points)
(I) because f (x) = 2cos2x − cos (2x + π)
2)=2cos2x+sin2x… (2 points)
=1+cos2x+sin2x… (4 points)
=
2sin(2x+π
4)+1… (6 points)
So f (π)
8)＝
2sin(π
4+π
4)+1＝
2+1… (7 points)
(II) because f (x) =
2sin(2x+π
4)+1
So t = 2 π
2＝π… (9 points)
The monotone decreasing interval of y = SiNx is (2k π + π)
2，2kπ+3π
2)，（k∈Z）… (10 points)
So let 2K π + π
2＜2x+π
4＜2kπ+3π
2… (11 points)
K π + π is obtained
8＜x＜kπ+5π
8… (12 points)
So the monotone decreasing interval of function f (x) is (K π + π)
8，kπ+5π
8)，（k∈Z）… (13 points)

What is the minimum positive period of the function f (x) = 2cos 2 x / 2 + cos (x + π / 3)?

f(x)=2(1+cosx)/2+cosxcosπ/3-sinxsinπ/3
=-(√3/2*sinx-3/2*cosx)+1
=-3/2*sin(x-π/3)+1
So t = 2 π / 2 = π

The function f (x) = cos2x − 2cos2x A monotone increasing interval of 2 is () A. (π 3，2π 3) B. (π 6，π 2) C. (0，π 3) D. (−π 6，π 6)

The function f (x) = cos2x − 2cos2x
2=cos2x-cosx-1，
The original function is regarded as G (T) = t2-t-1, t = cosx,
For G (T) = t2-t-1, when t ∈ [− 1, 1
2] G (T) is a minus function,
When t ∈ [1]
2,1], G (T) is an increasing function,
When x ∈ (π)
3，2π
3) T = cosx minus function,
And t ∈ (− 1)
2，1
2) The original function is monotonically increasing,
So choose a

A monotone increasing interval of the function y = cos ^ x-2cos ^ (x / 2)

Let t = cos (x / 2), - 1 1 / 2, monotonically increasing
cos(x/2) > 1/2,
-PI/3 < x/2 < PI/3,
-2PI/3 < x < 2PI/3.
So,
A monotone increasing interval of the function y = cos ^ x-2cos ^ (x / 2) is,
-2PI/3 < x < 2PI/3.

A monotone increasing interval of the function y = cos ^ 2x-2cos ^ 2 (x / 2)

y=cos^2x-2cos^2(x/2)
=cos^2x-cosx-1
=(cosx-1/2)^2-5/4
A monotone increasing interval [- π / 3,0]

Let f (x) = cos (2x Pai / 3) + cos2x - 1 If x belongs to [0, Pai / 2], find the maximum value of F (x) and the corresponding X

The minimum positive period is π, and the maximum value is root three-1 when x = π / 6

Given the function f (x) = sin (2x + π / 6) + sin (2x + π / 6) + cos2x, find the minimum positive period of function y = f (x) Speed! Explain! Process!