The function f (x) = If 3cos (3x − θ) − sin (3x − θ) is an odd function, then Tan θ is equal to () A. Three Three B. - Three Three C. Three D. - Three

The function f (x) = If 3cos (3x − θ) − sin (3x − θ) is an odd function, then Tan θ is equal to () A. Three Three B. - Three Three C. Three D. - Three

f(x)＝
3cos(3x−θ)−sin(3x−θ)＝−
3sin(3x−θ−π
3)，
If f (x) is an odd function, − θ π can be obtained
3 = k π, that is, θ = k π − π
3（k∈Z），
Therefore, Tan θ = Tan (K π − π)
3)＝tan(−π
3)＝−
3．
Therefore, D is selected

Function f (x) = radical 3cos (3x - θ) - sin (3x - θ) = 2Sin (3x - π / 3 - θ) Why?

This is the integration of trigonometric functions, which are available in high school textbooks. The required content of college entrance examination is to combine sine function and cosine function into a function. Asinx + bcosx = root sign (a ^ 2 + B ^ 2) * sin (x + θ), where Tan θ = B / a
Do you understand

Solving inequality: radical (x-1) - radical (2x-1) > 3x-2

And (2x) > - 1, and (2x) > - 1,
Now you can square directly: 3x-2-2 root sign [(x-1) (2x-1)] > 9x ^ 2-12x + 4,
The shift term: - 9x ^ 2 + 15x-6 > 2 root sign [(x-1) (2x-1)], and the left factorization result is: - 3 (3x-2) (x-1) > 2 radical sign [(x-1) (2x-1)],
Observe and find: on the right: constant greater than or equal to 0;
Left: (x-1) is greater than or equal to 0, 3x-2 > 0, so the left is always less than or equal to 0;
Therefore, the inequality is impossible to be established;
Therefore, the solution set of the original inequality is empty

Solve the inequality system {3x + 5} is greater than or equal to x-1,2x - radical sign 6

﹛3X+5≥X-1 2X-√6 ∵3X+5≥X-1 3X-X≥-1-5 2X≥-6 X≤-3 ∴2X≤-6 ∴2X-√6≤-6-√6

Inequality radical 2 (X-2)

2 (X-2) 0
So you can square both sides to get a quadratic inequality of one variable, as long as you solve the solution set of the quadratic inequality of one variable,
Combined with x > 0, we can get the correct answer
The square of 4 (X-2)

Solve the inequality group; (x + 3) > 0,2 (x + 1) + 3 ≥ 3x, and judge whether - 1 and radical 2 are the solutions of the inequality system

(x+3)>0
x>-3
2(x+1)+3>=3x
2x+2+3>=3x
3x-2x

Solve the inequality group x + 4 > 0,2 (x-1) + 3 ≥ 3x, and judge whether x = (√ 3-1) / 2 satisfies the modified inequality

x+4＞0,①
2﹙x-1﹚+3≥3x②
From ① to
x＞-4
From (2) to (2)
2x-2＋3≥3x
x≤1
Namely
-4＜x≤1
x=﹙√3-1﹚/2
0＜﹙√3-1﹚/2＜1
therefore
The inequality is satisfied

{x + 3 > 0 2 (x-1) + 3 ≥ 3x and judge whether x = √ 3 is the solution of the inequality system

If x + 3 > 0, then x > - 3
2 (x-1) + 3 ≥ 3x, then 2x-2 + 3 ≥ 3x, then 3x-2x ≤ 1, then x ≤ 1
So - 3 < x ≤ 1
Because √ 3 ＞ 1
So x = √ 3 is not the solution of the inequality system

Given that the final edge of angle α is on the line y = radical 3 * x / 3, then how much sin α is equal to and how much Tan α is equal to

Given that the final edge of the angle α is on the line y = radical 3 * x / 3, then sin α is equal to 1 / 2 or - 1 / 2, how much is tan α equal to √ 3 / 3

Given that the angle between the straight line y = radical 3x + 3 and the positive half axis of X axis is α, find the values of Tan α, sin α, cos α

The line y = radical 3x + 3 intersects with X axis and (- √ 3,0), intersects with y axis and (0,3),
So tan α = 3 / √ 3 = √ 3, so α = 60 degrees
So sin α = √ 3 / 2,
cos α=1/2