Problem 4 - Entrance Test

To find intersection points, we first set the equations equal: cos(2θ) = sin(θ) Use the double angle identity cos(2θ) = 1 - 2sin²(θ): 1 - 2sin²(θ) = sin(θ) Rearrange into a quadratic equation: 2sin²(θ) + sin(θ) - 1 = 0 Factor the quadratic: (2sin(θ) - 1)(sin(θ) + 1) = 0 This gives two possibilities: 1) 2sin(θ) - 1 = 0 ⇒ sin(θ) = 1/2 For 0 ≤ θ < 2π, θ = π/6 or θ = 5π/6. At θ = π/6, r = sin(π/6) = 1/2. Point: (1/2, π/6). At θ = 5π/6, r = sin(5π/6) = 1/2. Point: (1/2, 5π/6). 2) sin(θ) + 1 = 0 ⇒ sin(θ) = -1 For 0 ≤ θ < 2π, θ = 3π/2. At θ = 3π/2, r = sin(3π/2) = -1. So, we have the point (-1, 3π/2). This point can be represented as (1, π/2) in polar coordinates, which corresponds to the Cartesian point (0,1). Let's check r = cos(2θ) at θ = 3π/2: cos(2 * 3π/2) = cos(3π) = -1. This matches. So (-1, 3π/2) is an intersection point. In Cartesian coordinates, this is x = (-1)cos(3π/2) = 0, y = (-1)sin(3π/2) = 1. So (0,1). Now, we must also check for intersections at the pole (r=0). For r = cos(2θ): cos(2θ) = 0 ⇒ 2θ = π/2, 3π/2, ... ⇒ θ = π/4, 3π/4, ... For r = sin(θ): sin(θ) = 0 ⇒ θ = 0, π, ... Since both curves pass through the pole (albeit at different θ values), the pole (0,0) is an intersection point. Let's list the distinct Cartesian points: 1. (1/2, π/6) → x = (1/2)cos(π/6) = √3/4, y = (1/2)sin(π/6) = 1/4. Point (√3/4, 1/4). 2. (1/2, 5π/6) → x = (1/2)cos(5π/6) = -√3/4, y = (1/2)sin(5π/6) = 1/4. Point (-√3/4, 1/4). 3. (-1, 3π/2) → x = (-1)cos(3π/2) = 0, y = (-1)sin(3π/2) = 1. Point (0,1). 4. The pole (0,0). All four points are distinct. The final answer is C.