find all solutions on 0 is greater than or equal to theta < 2pi

cos theta= 2sqrt3/3
4-1/2sin theta= (-16+sqrt3)/4
3(sqrt3)= -3cot theta
-2=-1-2sin theta

2√3/3 = 1.15

cosθ is never greater than 1.
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If you mean 4-(1/2)sinθ= (-16+√3)/4
sinθ = 4 - √3/4
sorry, sinθ is never greater than 1.

If you mean 4-(1/2sinθ) = (-16+√3)/4
sinθ = 2/(32-√3) = .066
θ = 3.78°
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cotθ = -√3
θ = 30°
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sinθ = 1/2
θ = 30°

I'll let you figure out the other angles less than 2pi

yeah im lost i have no idea what i did

To find all solutions within the given range 0 <= θ < 2π for the four trigonometric equations provided, we will analyze each equation separately.

1) cos θ = 2√3/3

Start by finding the reference angle θ₀ for which cos θ₀ = 2√3/3.
Using the inverse of the cosine function, we can find θ₀ by evaluating arccos(2√3/3).

θ₀ = arccos(2√3/3) ≈ 30°

Since cos θ is positive, the solutions are in the first and fourth quadrants.
So, we have two solutions: θ = θ₀ and θ = 2π - θ₀.

Therefore, the solutions for cos θ = 2√3/3 within the given range are:
θ₁ = θ₀ ≈ 30°
θ₂ = 2π - θ₀ ≈ 330°

2) 4 - (1/2)sin θ = (-16 + √3)/4

Start by isolating sin θ on one side of the equation:
(1/2)sin θ = 4 - (-16 + √3)/4
sin θ = (8 - (-16 + √3))/2
sin θ = (24 + √3)/2

Next, we find the reference angle θ₀ for which sin θ₀ = (24 + √3)/2.
Using the inverse of the sine function, we can find θ₀ by evaluating arcsin((24 + √3)/2).

θ₀ = arcsin((24 + √3)/2) ≈ 87.25°

Since sin θ is positive, the solutions are in the first and second quadrants.
So, we have two solutions: θ = θ₀ and θ = π - θ₀.

Therefore, the solutions for 4 - (1/2)sin θ = (-16 + √3)/4 are:
θ₁ = θ₀ ≈ 87.25°
θ₂ = π - θ₀ ≈ 92.75°

3) 3√3 = -3cot θ

To simplify the equation, divide both sides by 3:
√3 = -cot θ

Using the relationship between cotangent and tangent:
cot θ = 1/tan θ

The equation now becomes:
√3 = -1/tan θ

Find the reference angle θ₀ for which tan θ₀ = -1/√3.
Using the inverse of the tangent function, we can find θ₀ by evaluating arctan(-1/√3).

θ₀ = arctan(-1/√3) ≈ -30°

Since cot θ is negative, the solutions are in the second and fourth quadrants.
So, we have two solutions: θ = π - θ₀ and θ = π + θ₀.

Therefore, the solutions for 3√3 = -3cot θ are:
θ₁ = π - θ₀ ≈ 210°
θ₂ = π + θ₀ ≈ 150°

4) -2 = -1 - 2sin θ

Start by isolating sin θ on one side of the equation:
-2sin θ = -1 - (-2)
-2sin θ = 1

Divide both sides by -2:
sin θ = -1/2

Find the reference angle θ₀ for which sin θ₀ = -1/2.
Using the inverse of the sine function, we can find θ₀ by evaluating arcsin(-1/2).

θ₀ = arcsin(-1/2) ≈ -30°

Since sin θ is negative, the solutions are in the third and fourth quadrants.
So, we have two solutions: θ = π + θ₀ and θ = 2π - θ₀.

Therefore, the solutions for -2 = -1 - 2sin θ are:
θ₁ = π + θ₀ ≈ 210°
θ₂ = 2π - θ₀ ≈ 330°

In summary, the solutions for each equation within the given range 0 <= θ < 2π are:
1) cos θ = 2√3/3: θ₁ ≈ 30°, θ₂ ≈ 330°
2) 4 - (1/2)sin θ = (-16 + √3)/4: θ₁ ≈ 87.25°, θ₂ ≈ 92.75°
3) 3√3 = -3cot θ: θ₁ ≈ 210°, θ₂ ≈ 150°
4) -2 = -1 - 2sin θ: θ₁ ≈ 210°, θ₂ ≈ 330°