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

To find all the solutions within the given range, we need to manipulate each equation separately to solve for theta.

1. cos(theta) = 2sqrt(3)/3:
Start by taking the inverse cosine (cos^(-1)) of both sides of the equation.
cos^(-1)(cos(theta)) = cos^(-1)(2sqrt(3)/3)

Since cos^(-1)(cos(theta)) gives us the principle value of theta (between 0 and pi), we need to consider the angle on the unit circle where cos(theta) = 2sqrt(3)/3. This angle is pi/6.

Therefore, one solution within the range 0 <= theta < 2pi is theta = pi/6.

2. 4 - (1/2)sin(theta) = (-16 + sqrt(3))/4:
Start by isolating sin(theta) by subtracting 4 from both sides:
(1/2)sin(theta) = (-16 + sqrt(3))/4 - 4

Next, simplify the right side of the equation:
(1/2)sin(theta) = (-16 + sqrt(3) - 16)/4
(1/2)sin(theta) = (-32 + sqrt(3))/4

Multiply both sides of the equation by 2 to eliminate the fraction:
sin(theta) = 2 * (-32 + sqrt(3))/4

Simplify further:
sin(theta) = -32 + sqrt(3)/2

Now, take the inverse sine (sin^(-1)) of both sides of the equation to find the principle value:
sin^(-1)(sin(theta)) = sin^(-1)(-32 + sqrt(3)/2)

The inverse sine function between the range -pi/2 to pi/2 gives us the principle value of theta.

However, since the given range is 0 <= theta < 2pi, we need to consider the quadrant where sin(theta) = -32 + sqrt(3)/2. This quadrant is the fourth quadrant (between 3pi/2 and 2pi).

Therefore, one solution within the range 0 <= theta < 2pi is theta = 3pi/2 + sin^(-1)(-32 + sqrt(3)/2).

3. 3sqrt(3) = -3cot(theta):
Start by dividing both sides of the equation by 3:
sqrt(3) = -cot(theta)

To find theta, we need to take the inverse cotangent (cot^(-1)) of both sides of the equation:
cot^(-1)(sqrt(3)) = cot^(-1)(-cot(theta))

The cot^(-1) function gives us the principle value of theta (between 0 and pi), so we need to consider the angle on the unit circle where cot(theta) = -sqrt(3). This angle is 5pi/6.

Therefore, one solution within the range 0 <= theta < 2pi is theta = 5pi/6.

4. -2 = -1 - 2sin(theta):
Start by isolating sin(theta) by adding 1 to both sides:
-1 + 2 + 1 = -2 + 2sin(theta) + 1
1 = 2sin(theta)

Divide both sides of the equation by 2 to solve for sin(theta):
sin(theta) = 1/2

To find theta, take the inverse sine (sin^(-1)) of both sides of the equation:
sin^(-1)(sin(theta)) = sin^(-1)(1/2)

The inverse sine function between the range -pi/2 to pi/2 gives us the principle value of theta.

Since sin(theta) = 1/2, we need to consider the angles on the unit circle where sin(theta) = 1/2. These angles are pi/6 and 5pi/6 (first and second quadrants).

Therefore, additional solutions within the range 0 <= theta < 2pi are theta = pi/6 and theta = 5pi/6.

In summary, the solutions within the range 0 <= theta < 2pi are theta = pi/6, 3pi/2 + sin^(-1)(-32 + sqrt(3)/2), 5pi/6, and any other angles that are equivalent to these solutions by adding or subtracting multiples of 2pi.