Known function f (x) = 5 root 3cos square x + root 3sin square x-4sinxcosx (1) When x belongs to real genus, find the minimum value of F (x) (2) If π / 4 ≤ x ≤ 7 π / 24, find the monotone interval of F (x)

F (x) = radical 3 + 4 radical 3cos ^ 2x-2sin2x = radical 3 + 2 radical 3cos2x-2sin2x + 2 radical 3 = 3 radical 3 + 2 radical 3cos2x-2sin2x = 3 radical 3 + 4 [cos2xsin (π / 3) - cos (π / 3) sin2x] = 3 radical 3 + 4cos (2x + π / 3)
The minimum value is 3-4 π / 4 ≤ x ≤ 7 π / 24 π / 2

Find the function f (x) = 5 root 3cos ^ 2x radical 3sin ^ 2x-4sinxcosx (Wu / 4) It is preceded by F (x) = 5 root 3cos ^ 2x plus Radix 3sin ^ 2x minus 4sinxcosx

(sqrt is the square root) f (x) = 5sqrt (3) [cos (2x)] ^ 2 + sqrt (3) [sin (2x)] ^ 2 - 4sin (x) cos (x), because [cos (2x)] ^ 2 + [sin (2x)] ^ 2 = 1, and 2Sin (x) cos (x) = sin (2x), so = 5sqrt (3) {1 - [sin (2x)] ^ 2} + sqrt (3) [sin (2x)] ^ 2 - 2Sin (2x

Given the function f (x) = 3sin square x + 2 root sign 3sinxcosx-3cos square X,

F (x) = 3sin square x + 2 root sign 3sinxcosx-3cos square x
=Radical 3sin2x-3cos2x
=2 radical 3sin (2x - π / 3)

If the maximum value of the function f (x) = three times the root of sin2x + 2cosx + m on the interval [0, π / 2] is 6, find the value of M and the maximum value of this function when x belongs to R

f（x）=√3 sin⁡2x+2(cos⁡x )^2+m
=√3 sin⁡2x+cos⁡2x+m+1
=2sin⁡〖（2x+π/6）〗+（m+1）
Because f (x)_ Max = 6, so m = 3, f (x)_ min=2

Given that the sum of the maximum and minimum values of FX on the interval [- π / 6, π / 6] is 3 under the square + root sign of the function FX = 2cosx

f(x)=2cosx^2+√3sin2x+a
=cos2x+1+√3sin2x+a
=2(1/2cos2x+√3/2sin2x)+1+a
=2sin(2x+π/6)+1+a
F (x) in [- π / 6, π / 6] has: - π / 6 ≤ 2x + π / 6 ≤ π / 2
So we can get: when 2x + π / 6 = π / 2, the maximum value is 3 + a
When 2x + π / 6 = - π / 6, the minimum value is: a
Then we have: 3 + A + a = 3, so we can get: a = 0

Let y = cos square x + root sign 3sinxcosx + 1 / 2 (x ∈ R) determine the minimum positive period and simple decreasing interval of this function Find the maximum value of this function, and find the set of X when the maximum value of Y area

Y = square X of COS + root sign 3sinxcosx + 1 / 2
=1/2(1+Cos2x)+√3/2Sin2x+1/2
=Sin(π/6+2x)+1
When sin (π / 6 + 2x) = 1, that is, x = π / 6 + K π, K ∈ Z, y has a maximum value of 2
To sum up, y has a maximum value of 2, and the set of X is {x | x = π / 6 + K π, K ∈ Z}

The known function f (x) = 3sinxcosx−cos2x+1 2(x∈R)． (1) Find the minimum positive period of function f (x); (2) Find the function f (x) in the interval [0, π 4] The value range of the function value on

(1) Because f (x) = 0
Three
2sin2x−1
2cos2x… (4 points)
=sin(2x−π
6)… (6 points)
So the minimum positive period of F (x) is π (8 points)
(2) When x ∈ [0, π
4] 2 x − π
6∈[−π
6，π
3]… (10 points)
∴sin(2x−π
6)∈[−1
2，
Three
2]
Therefore, the value range obtained is [− 1
2，
Three
2]… (14 points)

The function f (x) = 2cosx squared + 2 Radix 3sinxcosx + 1 (1) If x belongs to [O, π], f (x) = a has two different roots. Find the value range of a and the sum of the two (2) The function y = f (x), X belongs to [π / 6,7 π / 6], what is the area of the graph surrounded by the line y = 4?

f(x)=2cos²x+2√3•sinxcosx+1=1+cos2x+√3•sin2x+1=2sin(2x+π/6)+2.
(1) If x ∈ [0, π], then (2x + π / 6) ∈ [π / 6,2 π + π / 6], the value range of F (x) is [0,4],
From the graph, if and only if a ∈ (0,3) ∪ (3,4), f (x) = a has exactly two distinct real roots, (when a = 3, there are three distinct real roots)
Therefore, the value range of a is (0,3) ∪ (3,4)
When a ∈ (0,3), the sum of the two is 4 π / 3; when a ∈ (3,4), the sum of the two is π / 3
(2) It can be seen from the graph that the area of the graph formed by the image of the function y = f (x), X ∈ [π / 6,7 π / 6] and the straight line y = 4 is 2 π
It is suggested that we can solve the problem by integral, or we can solve it by the complement method. We can separate the figure under the line y = 2 along the symmetry axis X = 2 π / 3, and respectively fill it above the line y = 2 and below the image of y = f (x), X ∈ [π / 6,7 π / 6], and then we can make the original graph into a rectangle with π and width of 2

The square root of F (x) = 6 * (cosx) 3 * sin2x 1) If a is an acute angle, f (a) = 3-2, root sign 3, find tan4a / 5

The square root sign of F (x) = 6 * (cosx) 3 * sin2x
=3cos2x+3-√3sin2x
=2√3sin(60-2x)+3
F (a) = 3-2 root sign 3 = 2 √ 3sin (60-2a) + 3
sin(60-2a)=-1
60-2a=-90,a=75
tan(4a/5)=tan60=√3

F (x) = - 2 root sign 3sin square x + sin2x + radical 3 find the minimum positive period and minimum value

y=-2√3（1-cos2x)/2+sin2x+√3
= -√3+√3cos2x+sin2x+√3
=2sin(2x+π/3)
So the minimum positive period π min - 2

Known function f (x) = 5 root 3cos square x + root 3sin square x-4sinxcosx (1) When x belongs to real genus, find the minimum value of F (x) (2) If π / 4 ≤ x ≤ 7 π / 24, find the monotone interval of F (x)