The equation SiNx radical 3cosx = 4m-6 / 4-m is solvable, and the value range of M is found

The equation SiNx radical 3cosx = 4m-6 / 4-m is solvable, and the value range of M is found

SiNx radical 3cosx = 4m-6 / 4-m
2sin(x-π/3)=(4m-6)/(4-m)
-2≤(4m-6)/(4-m)≤2
(4m-6)/(4-m)≥-2
(4m-6+8-2m)(m-4)≤0
(2m+2)(m-4)≤0
-1≤m<4
(4m-6)/(4-m)≤2
(4m-6-8+2m)(m-4)≥0
(6m-14)(m-4)≥0
M ≤ 7 / 3 or M > 4
Take the intersection to get - 1 ≤ m ≤ 7 / 3

SiNx radical 3cosx = 6-4m, in order to make it meaningful, find the value range of M

SiNx radical 3cosx = 6-4m
2(sinxcosπ/3-cosxsinπ/3) = 6-4m
2sin(x-π/3) = 6-4m
-2≤2sin(x-π/3)≤2
∴-2≤6-4m≤2
-8≤-4m≤4
-1≤m≤2

It is known that the equation SiNx + √ 3cosx = m has two distinct real roots a and B in the open interval (0,2 Π). Find the value range of real root m and the value of a + B

SiNx + radical 3cosx = 2Sin (x + π / 4) because x belongs to (0,2 π), so x + π / 4 belongs to (π / 4,9 π / 4), so - 2 ≤ 2Sin (x + π / 4) ≤ 2, so - 2 ≤ m ≤ 2 can be obtained according to the sine function, if x is in (0,2 π), then X1 + x2 = 2 π, so a + π / 4 + B + π / 4 = 2 π, so a + B = 3 π / 2

Equation SiNx + √ 3cosx + a = there are two different real roots a in the interval [0, π / 2], B.1 find the value range of real number a (2) find the value of a + B

In the interval [0, π / 2], there are two different real root a, a, B2 (sinxcos (π / 3) + sin (π / 3 + 3 + x) + a = 0sin (x + π / 3 / 3) + a = 0sin (x + π / 3) = - A / 2 π / 3 ≤ x + π / 3 ≤ 5 π / 3 ≤ 5 π / 6-2 ≤ a ≤ - 3 ^ (12 /) sin (a + π / 3) = sin (π - A - π / 3) = sin (π / 3-A + 3 + π / 3) π / 3 / 3) π / 3 (3) π / 3) π / 3 (3) π / 3) π / 3) π / 3) π / 3) π / 3 3-A = Ba + B = π / 3

Let the equation SiNx + √ 3cosx = a have two distinct real roots X1 and X2 in the interval (0,2 π). Find the value range of a and the value of X1 + x2

sinx+√3cosx=a
sinx*1/2+√3cosx/2=a/2
sin(x+π/3)=a/2
When-2

Function f (x) = SiNx- The monotone increasing interval of 3cosx (x ∈ [- π, 0]) is () A. [-π，-5 6π] B. [-5 6π，-π 6] C. [-π 3，0] D. [-π 6，0]

f（x）=sin x-
3cos x=2sin（x-π
3），
Because X - π
3∈[-4
3π，-π
3]，
So x - π
3∈[-1
2π，-π
3]，
We get x ∈ [- π
6，0]，
Therefore, D is selected

The function y = SiNx+ 3cosx is in the interval [0, π 2] The value range on is______ .

y＝sinx+
3cosx=2sin（ x+π
3）
∵x∈[0，π
2]
∴x+π
3∈[π
3，5π
6]
∴1
2≤sin（ x+π
3）≤1
∴1≤y≤2
So the answer is: [1,2]

Function f (x) = SiNx- The monotone increasing interval of 3cosx (x ∈ [- π, 0]) is () A. [-π，-5 6π] B. [-5 6π，-π 6] C. [-π 3，0] D. [-π 6，0]

f（x）=sin x-
3cos x=2sin（x-π
3），
Because X - π
3∈[-4
3π，-π
3]，
So x - π
3∈[-1
2π，-π
3]，
We get x ∈ [- π
6，0]，
Therefore, D is selected

When - 2 parts π ≤ x ≤ 2 parts π, the value range of function y = SiNx + root 3cosx

y=sinx+√ 3cosx
y=2(1/2sinx+√ 3/2cosx)
y=2sin(x+π/3)
-π/2≤x≤π/2
-π/6≤x+π/3≤5π/6
The value range is [- 1 / 2, 1]

Given the function, y = root 3cosx SiNx, (1) y > 0, find the value range of X

Trigonometric function of an angle:
y=2cos(π/6+x)
From Y > 0
The results are as follows:
-π/2+2kπ