Solving the inequality Tan (2x + π / 3) > (radical 3) / 3

Solving the inequality Tan (2x + π / 3) > (radical 3) / 3

tan(π/6)=√3/3
The tan period is π
So tan (K π + π / 6) = √ 3 / 3
Tan is an increasing function in a period (K π - π / 2, K π + π / 2)
tan(2x+π/3)>tan(kπ+π/6)
So K π + π / 6 < 2x + π / 3K π + π / 6 - π / 3 < 2x + π / 3 - π / 3K π - π / 6 < 2x, so K π / 2 - π / 12
Homework help users 2017-10-03
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How many degrees of Tan is equal to 2 + radical 3

Tan75 = Tan (45 + 30) = (tan45 + tan30) / (tan45-tan30) = (1-radical 3 / 3) / (1 + radical 3 / 3) = 2 + radical 3
(Note: the degree is omitted in each corner)

Straight line 3x+ The inclination angle of 3y-1 = 0 is 0______ .

Because of the straight line 3x+
The slope of 3y-1 = 0 is − 3
3=-
3. If the inclination angle is θ, then Tan θ=-
3，θ=120°，
So the answer is: 120 degrees

The inclination angle of the root of the line 3x + 3y-3 = 0 is

Let the inclination angle be a, then Tana = K (slope)
∵3x+3y-3=0
∴k＝3／（－3）＝－1
∴tana＝－1
∴a＝3π／4
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The angle between x-radical 3Y + 2 = 0 and Radix 3x + 3y-4 = 0

The first slope is √ 3 / 3 and the second slope is - √ 3 / 3. So the angle is 60 degrees

The inclination angle of 3x + radical 3Y + 1 = 0

Because 3x + √ 3Y + 1 = 0
therefore
y=-√3x-√3/3
So k = - √ 3
Tana = - √ 3
Then the inclination angle a = 120 degrees

Real numbers x, y, m satisfy the following conditions: 3x+5y−2−m+ 2x+3y−m＝ x−199+y• 199 − x − y, then M = () A. 100 B. 200 C. 201 D. 2001

According to the meaning of the title
x−199+y≥0
199 − x − y ≥ 0, x + y = 199,
Qi
3x+5y−2−m+
2x+3y−m=0，
Qi
x+y＝199
3x+5y−2−m＝0
2x+3y−m＝0 ，
By solving the equation
x＝396
y＝−197
m＝201 ．
Therefore, C

Is 3x ^ 2-radical x = 0 a quadratic equation of one variable

3x ^ 2-radical x = 0 is not a quadratic equation of one variable
3x ^ 2-radical x = 0 is an irrational equation

The univariate quadratic equation with negative root 2 plus root 3 and negative root 2 minus root 3 as roots is______ You want to know how,

Let the quadratic equation of one variable be x 2 - MX + n = 0, and obtain X1 = - √ 2 + √ 3; x2 = - √ 2 - √ 3M = X1 + x2 = - 2 √ 2, n = x1 · x2 = - 1, then the quadratic equation of one variable is x? + 2 √ 2x-1 = 0. If multiple groups are needed, we only need to multiply a coefficient which is not 0 by the integral formula

The function f (x) = 2sinx-1-a in X ∈ [π If there are two zeros on π], then the value range of real number a is () A. [-1，1] B. [0， 3−1] C. [0，1） D. [ 3−1，1)

∵ when x ∈ [π]
3, π], t = SiNx in the interval (π)
3，π
2) It is an increasing function,
In the interval (π
2, π), and sin π
3=sin2π
Three
When x ∈ [π]
3，2π
3] And X ≠π
When 2, there are two independent variables X corresponding to the same SiNx
That is, if t ∈[
Three
If there are two zeros of sin T, x = 1
∵ f (x) = 2sinx-1-a in X ∈ [π]
There are two zeros on 3, π], that is, 1 + a
2 = SiNx in X ∈ [π
There are two zeros on 3, π],
∴1+a
2∈[
Three
2,1), and a ∈[
3−1，1)
Therefore, D