If
(2 sin(x) - 1) (2 cos²(x) - 1) = 0
then either
2 sin(x) - 1 = 0 or 2 cos²(x) - 1 = 0
sin(x) = 1/2 or cos(2x) = 0
The first equation gives two families of solutions,
x = arcsin(1/2) + 2nπ = π/6 + 2nπ
x = π - arcsin(1/2) + 2nπ = 5π/6 + 2nπ
where n is any integer; we get the solutions x = π/6 and x = 5π/6 in the interval 0 ≤ x < 2π when n = 0.
The second equation also gives two families of solutions,
2x = arccos(0) + 2nπ = π/2 + 2nπ
2x = -arccos(0) + 2nπ = -π/2 + 2nπ
Divide both sides by 2 :
x = π/4 + nπ
x = -π/4 + nπ
Then the solutions we want are x = π/4 and x = 5π/4 (when n = 0 and 1 in the first family, respectively) and x = 3π/4 and x = 7π/4 (when n = 1 and 2 in the second family).
We have 6 solutions, so A is the correct choice.
