Final answer:
The points of intersection between r = 2 sin(2θ) and r = 1, ordered from smallest to largest θ, are π/12, 5π/12, 13π/12, and 17π/12.
Step-by-step explanation:
To find all points of intersection of the curves r = 2 sin(2θ) and r = 1 within the range 0 ≤ θ ≤ 2π and r ≥ 0, we can set the two equations equal to each other because at the intersection points, both r values must be the same:
1 = 2 sin(2θ)
Now, solve for θ:
sin(2θ) = ½
Let's find the angles that satisfy this equation within the given range. Using the unit circle and inverse trigonometric functions:
2θ = π/6, 5π/6
Divide both sides by 2 to solve for θ:
θ = π/12, 5π/12
However, these are not the only solutions. We need to consider the periodicity of the sine function and find all solutions within 0 ≤ θ ≤ 2π. The general form for solutions to sin(2θ) = ½ is θ = nπ ± π/12 for n being an integer. After verifying all possible values, we confirm that the intersections occur at:
- θ = π/12
- θ = 5π/12
- θ = 13π/12
- θ = 17π/12
Since the question assumes r ≥ 0, any solution with a negative r can be discarded. Additionally, a point of intersection occurs at the pole when r = 0 for any angle θ; however, r = 0 does not occur for any angle θ given the equations being analyzed.
Therefore, the points of intersection ordered from smallest to largest θ are π/12, 5π/12, 13π/12, and 17π/12.