The solution set of inequality Tan (PI / 6-2x) greater than or equal to root 3 is

The solution set of inequality Tan (PI / 6-2x) greater than or equal to root 3 is

tan(pi/6-2x)=-tan（2x-pi/6)>=√3
That is, Tan (2x pi / 6) < = - √ 3
That is - pi / 2 < 2x pi / 6 < = - pi / 3 + k * PI, K belongs to integer
So the solution set of the original inequality is
(-pi/6+k*pi/2,pi/12+k*pi/2]

If sin (6 / 6-A) = 1 / 3, then cos (2 / 3 + 2a) is equal to 1 / 3

cos(2π/3＋2a)=－cos(π/3－2a)=－[cos2(π/6－a)]=－[1－2sin²(π/6－a)]=－[1－(2/9)]=－7/9

If cos (six fifths alpha) = one half, what is sin (AFA plus three) and COS (two thirds plus two Afars) Hurry up!

Cos = half,
cos（π/6-α）=1/2,
According to the induction formula, sin [π / 2 - (π / 6 - α)] = 1 / 2,
In other words, sin / 1 / π = 1,
cos（2π/3+2α）= cos[2(π/3+α)]
=1-2 sin²(π/3+α) =1/2.

Given sin (6 / 5 + R) = 3 / 5, what is the value of COS (3 / 3-R)?

The relation of trigonometric function: Sina = cos (л / 2-A),
So sin (л / 6 + a) = cos [л / 2 - (л / 6 + a)]
=cos(л/3-a)
=3/5

Sin (3 π / 2-2x) = 3 / 5 to find tan? X

According to the induction formula, sin (3 π / 2-2x) = - cos2x = 3 / 5cos2x = 2cos? X-1 = - 3 / 52cos? X = 2 / 5cos? X = 1 / 5sin? X = 1-cos? X = 4 / 5, so tan? X = sin? X / cos? X = 4 / 5 / 1 / 5 = 4

Sin (π - α) = - 2 / 3 α∈ (- π / 2,0), then Tan α is equal to

sin(π-α)=sinα=-2/3
α∈（-π/2,0）
So cos α > 0
sin²α+cos²α=1
So cos α = √ 5 / 3
tanα=sinα/cosα=-2√5/5

If sin (α + β) = 1 / 5, sin (α - β) = 3 / 5, then what is tan α / Tan β equal to?

sin（α+β)=sinαcosβ+cosαsinβ=1/5 ,sin(α-β)=sinαcosβ-cosαsinβ=3/5,sin（α+β)+sin（α-β)=4/5 ,2sinαcosβ=4/5 ,sinαcosβ=2/5sin（α+β)-sin（α-β)=-2/5 ,2cosαsinβ=-2/5 ,cosαsinβ-1/5ta...

If f (x) = sin (2x + φ) + root three cos (2x + θ) is an odd function, then Tan θ is equal to

The definition domain of the function f (x) = sin (2x + φ) + root three cos (2x + θ) = 2Sin (2x + θ + π / 3) is r
Then f (0) = 0
So: 0 = sin (θ + π / 3),
=>θ+π/3=2kπ ,k∈Z
That is, θ = 2K π - π / 3
So:
tanθ=tan(2kπ-π/3)=-tanπ/3=-√3

Find cos (120-2x) with sin (x + 30) = 1 / 3

cos(2x+60)
=cos[2(x+30)]
=1-2sin²(x+30)
=7/9
cos(120-2x)
=cos[180-(2x+60)]
=-cos(2x+60)
=-7/9

Sin (30. + x) = 1 / 3, cos (120. - 2x) value is?

sin(30º + x) = 1/3
cos(120º - 2x)
= - cos[180º -(120º - 2x)]
= - cos(60º + 2x)
= - cos[2(30º + x)]
= 2sin²(30º + x) - 1
= -7/9