Final answer:
The solutions to the equation 2sin²(2x)cos(3x) - cos(3x) = 0 on the interval [0, π) are x = π/6 and x = π/2. These solutions are found by setting cos(3x) and 2sin²(2x) - 1 each to zero and solving for x.
Step-by-step explanation:
To find all solutions on the interval [0, π) for the equation 2sin²(2x)cos(3x) - cos(3x) = 0, we factor out cos(3x) from both terms, which gives us cos(3x)(2sin²(2x) - 1) = 0. This results in two separate equations, cos(3x) = 0 and 2sin²(2x) - 1 = 0. The first equation, cos(3x) = 0, has solutions for x when 3x is an odd multiple of π/2, which in the interval [0, π) means 3x = π/2 or 3x = 3π/2; that is, x = π/6 or x = π/2. The second equation 2sin²(2x) - 1 = 0 can be rewritten as sin²(2x) = 1/2, and since sin(2x) can be positive or negative, we have sin(2x) = ±√(1/2). However, in the interval [0, π), sin(2x) is non-negative, and so we only consider the positive root sin(2x) = √(1/2), which corresponds to 2x = π/4 or 2x = 3π/4; that is, x = π/8 or x = 3π/8, neither of which are options provided. After analyzing both equations, the correct solutions from the provided options are x = π/6 (a) and x = π/2 (b).