If the function y = A and the function y = SiNx + root sign 3cosx have two different intersections in X ∈ (0,2 π), then the value range of real number a is?

Therefore, the value of cosin + 2x is in the range of 2x = 2x

The function f (x) = SiNx + root 3cosx, X belongs to the value range of [0,5 / 12x]

F (x) = SiNx + radical 3cosx
=2*(sinx*cosπ/3+cosx*sinπ/3)
=2sin(x+π/3)
Because 0 < = x < = 5 / 12 π, π / 3 < = x + π / 3 < = 3 / 4 π
Sin (x + π / 3) belongs to [2 ^ (- 1 / 2), 1]
F (x) = 2Sin (x + π / 3) belongs to [2 ^ (1 / 2), 2]
So the value range is [2 ^ (1 / 2), 2]

Given the vector a = (root 3cosx, 0), B = (0, SiNx), note that the function f (x) = (a + b) squared + Radix 3sin2x, X belongs to the [4-th, 2-th] school. Find the maximum and minimum values of the function f (x)

I checked it. I'm tired of typing I went to drink milk tea
Answer: 2 + √ 3
Process:
a+b=(√3cosx,sinx)
F (x) = 3cosx squared + SiNx squared + √ 3sin2x
=1 + 2cosx square + √ 3sin2x
=2+cos2x+√3sin2x
=2 + 2cos (2x Pai / 3)
X belongs to [Pai / 4, Pai / 2]
2X Pai / 3 belongs to [Pai / 6,2 Pai / 3]
Cos (2x Pai / 3) belongs to [- 1 / 2, √ 3 / 2]
Therefore, the maximum value is obtained at x = Pie / 4, which is 2 + √ 3

Given the vector a = (root 3cosx, 0), vector b = (0, SiNx), Let f (x) = (a + b) ^ 2 + root 3sin2x. Find the monotone increasing interval of (1) function f (x) Given vector a = (root 3cosx, 0), vector b = (0, SiNx), Let f (x) = (a + b) ^ 2 + Radix 3sin2x. Find the minimum value of (1) function f (x) and * * (2) monotone increasing interval of function f (x)

The minimum value of function f (x) is 1

If f (x) = SiNx + 3cosx, X ∈ R, then the range of F (x) is () A. [1，3] B. [1，2] C. [− 10， 10] D. [0， 10]

f（x）=sinx+3cosx=
10sin（x+φ）
∵x∈R，
The value range of F (x) is[-
10，
10]
Therefore, C

The maximum and minimum values of y = SiNx + cosx + 1 / Radix (1 + | sin2x |)

1+|sin2x|=|sinx|^2+2*|sinx|*|cosx|+|cos|^2=(|sinx|+|cos|)^2
√(1+|sin2x|)=|sinx|+|cosx|
One

The maximum value of y = sinx-3 / cosx + 4 is

y=sinx-3/cosx+4
ycosx+4y=sinx-3
sinx-ycosx=4y+3
√(1+y²)sin(x-∅)=4y+3
sin(x-∅)=(4y+3)/√(1+y²)
So | (4Y + 3) / √ (1 + y?)
|4y+3|≤√(1+y²)
Square on both sides
16y²+24y+9≤1+y²
15y²+24y+8≤0
(-12-2√6)/15≤y≤(-12+2√6)/15
The maximum value is (- 12 + 2 √ 6) / 15

The maximum value of y = SiNx ^ 3 + cosx ^ 3 on [- π / 4, π / 4]

y＝sinx^3+cosx^3=(sinx + cosx )(sinx^2 - sinxcosx + cos^2)=(sinx + cosx)(1 - sin(2x)/2 )
=It's easy to do it yourself

How does | (3 + 4Y) / √ (1 + y |) ≤ 1 in the process of finding the maximum value of y = (sinx-3) / cosx + 4)

y=sinx-3/cosx+4
ycosx+4y=sinx-3
sinx-ycosx=4y+3
√(1+y²)sin(x-∅)=4y+3
sin(x-∅)=(4y+3)/√(1+y²)
On the left side of the above formula is a sine function whose range is [- 1,1]
So | (4Y + 3) / √ (1 + y?)

What is the maximum value of y = (cosx-3) / (SiNx + 4)

Let k = y
Then K is the slope of the line passing through two points a (SiNx, cosx) and B (- 4,3)
sin²x+cos²x=1
So a is on the unit circle
So the line AB and the unit circle have a common point
The line passing through B is Y-3 = K (x + 3)
kx-y+3+3k=0
If the line AB and the unit circle have a common point, the radius from the center of the circle to the straight line is less than or equal to the radius
So | 0-0 + 3 + 3K | / √ (K | + 1) ≤ 1
square
9k²+18k+9≤k²+1
4k²+9k+4≤0
(-9-√17)/8

If the function y = A and the function y = SiNx + root sign 3cosx have two different intersections in X ∈ (0,2 π), then the value range of real number a is?