find to the nearest minute, all positive values in the interval 0° ≤ θ < 180° that satisfy the equation 2 tan^2 θ - tan θ = 3

Substitute t=tan(&theta)

and rewrite the equation as:
2t²-t-3=0
which factorizes to
(2t-3)(t+1)=0
Solve for values of t, and find the solutions for θ.
Note: 1 degree = 60 minutes.

To find the values of θ that satisfy the equation 2 tan^2 θ - tan θ = 3 within the given interval, we will solve the equation step by step:

Step 1: Substitute tan θ = x
The equation becomes: 2x^2 - x = 3

Step 2: Rearrange the equation:
2x^2 - x - 3 = 0

Step 3: Solve the quadratic equation:
To solve the quadratic equation, we can either factorize it or use the quadratic formula. In this case, factoring is not straightforward, so we will use the quadratic formula.

The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a), where a = 2, b = -1, and c = -3.

Step 4: Substitute the values into the quadratic formula:
x = (-(-1) ± √((-1)^2 - 4(2)(-3))) / (2(2))
x = (1 ± √(1 + 24)) / 4
x = (1 ± √25) / 4
x = (1 ± 5) / 4

Step 5: Simplify fractions:
x1 = (1 + 5) / 4 = 6 / 4 = 3 / 2
x2 = (1 - 5) / 4 = -4 / 4 = -1

Step 6: Substitute tan θ = x back into the equation:
For x = 3 / 2:
tan θ = 3 / 2
θ = arctan(3 / 2)

For x = -1:
tan θ = -1
θ = arctan(-1)

Step 7: Convert the angle values to degrees.
θ = arctan(3 / 2) ≈ 56.31°
θ = arctan(-1) ≈ -45°

Step 8: Find all positive values of θ in the interval 0° ≤ θ < 180°.
Since 56.31° and -45° are both within this interval, we only consider the positive values.

Therefore, the solutions to the equation 2 tan^2 θ - tan θ = 3 within the given interval are approximately:
θ1 ≈ 56.31°
θ2 ≈ 135°

To summarize, the two positive values of θ to the nearest minute that satisfy the equation are approximately 56 minutes and 135 minutes.