Solution set of inequality 2x-1 / 3x + 1 > 1 The numerator is 2x-1 and the denominator is 3x + 1 What is the solution set

Solution set of inequality 2x-1 / 3x + 1 > 1 The numerator is 2x-1 and the denominator is 3x + 1 What is the solution set

It is easy to discuss by classification
When 3x + 1 > 0, that is, x > 1 / 3, 2x-1 > 3x + 1

If a = {(x, y) x + 3Y = 7, B = {(x, y) x-3y = 1}, then a ∩ B

A = {(x, y) x + 3Y = 7, B = {(x, y) x-3y = 1}
x+3y=7,x-3y=1
x = 4,y = 1
therefore
A∩B={(4,1)}

If a = {2,4, x ^ 3-2x ^ 2-x + 7}, B = {- 4, y + 3, y ^ 2-2y + 2, y ^ 3 + y ^ 2 + 3Y + 7} and a ∩ B = {2,5}, then a ∪ B=
How did a ∪ B = {- 4,2,4,5,25} 25 get it?

Let y ＾ 2-2y + 2 = 2, solve y = 0 or y = 2, substituting y = 0 into B does not conform to the meaning of the problem, substituting y = 2 into B, we get b = {- 4,5,2,25}

The known set a = {x | x + 2Y = 4}, B = {(x, y) | x-3y = - 1}, a ∩ B =?

x+2y=4,x-3y=-1,
By subtracting the two formulas, we get 5Y = 5, that is, y = 1
Substituting y = 1 into x + 2Y = 4, we get x + 2 = 4, and the solution is x = 2,
So a ∩ B = {(2,1)}

Why does sin (90 degree + 2x) / 1 + cos (90 degree + 2x) get Tan (45 degree + x)

Left side = cos2x / (1-sin2x) = (cosx ^ 2-sin2x ^ 2) / (SiNx ^ 2 + cosx ^ 2-2sinx * cosx)
=(cosx + SiNx) × (cosx SiNx) / (cosx SiNx) ^ 2
=(cosx + SiNx) / (cosx SiNx) divided by cosx
=(1 + TaNx) / (1-tanx) = Tan (45 degrees + x)

Given SiNx = - 1 / 3, X ∈ (π, 3 π / 2), the process of finding sin (2x-45) and COS (2x + 45) is given

I won't write about the process, I'll tell you the idea
According to the two sum difference formulas and double angle formula, all are marked with SiNx and cosx
If we know the range of SiNx and X, we can get cosx

Given sin ^ 2x + SiNx = 1, find the value of COS ^ 4x = cos ^ 2x

sin^2X+sinX=1,sinX=1-sin^2X=cos^2X cos^4X=sin^2X
Cos ^ 4x = cos ^ 2x should be +?
cos^4+cos^2X=sin^2X+cos^2X=1

F (x) = (COS ^ 2x-sin ^ 2x) / 2

f(x)=1/2cos2x
G (x) up translation 1 / 4 -- "1 / 2sin2x"
Shift π / 4 to the left

sin（π/6+x）=1/3 cos（2π/3-2x）=?

cos(π/3-x)
=sin[π/2-(π/3-x)]
=sin(π/6+x)
=1/3
So cos (2 π / 3-2x)
=cos[2(π/3-x)]
=2cos²(π/3-x)-1
=-7/9

1/2[cos(2x)cos(π/6)+sin(2x)sin(π/6)]-1/2[cos(2x)cos(π/6)-sin(2x)sin(π/6)]=1/2sin(2x)
How to get 1 / 2Sin (2x)?

The result of 1 / 2 subtraction is 2Sin (2x) sin (π / 3)
sin(π/3)=1/2
So the result is 1 / 2Sin (2x)