Final answer:
The quadratic equation in cos(theta) can be factored and solved for theta within the range of 0 to less than 2π, yielding solutions for theta at 0, π/3, and 5π/3.
Step-by-step explanation:
The equation 2cos²(theta) - 3cos(theta) + 1 = 0 can be solved by considering it as a quadratic equation in terms of cos(theta). Let's set u = cos(theta), this transforms the equation to 2u² - 3u + 1 = 0. We can factor this quadratic to (2u - 1)(u - 1) = 0, leading to u = 1/2 or u = 1. Since u is the cosine of theta, we now solve for the angles within the given range 0 ≤ θ < 2π. For u = 1/2, the angles are theta = π/3 and 5π/3. For u = 1, the angle is theta = 0. So, the solutions for theta are 0, π/3, and 5π/3.