(i) 3 csc²(x) - 4 = 0
3 csc²(x) = 4
csc²(x) = 4/3
sin²(x) = 3/4
sin(x) = ± √3/2
x = arcsin(√3/2) + 2nπ or x = arcsin(-√3/2) + 2nπ
x = π/3 + 2nπ or x = -π/3 + 2nπ
where n is any integer. The general result follows from the fact that sin(x) is 2π-periodic.
In the interval 0 ≤ x ≤ 2π, the first family of solutions gives x = π/3 and x = 4π/3 for n = 0 and n = 1, respectively; the second family gives x = 2π/3 and x = 5π/3 for n = 1 and n = 2.
(ii) 4 cos²(x) + 2 cos(x) - 2 = 0
2 cos²(x) + cos(x) - 1 = 0
(2 cos(x) - 1) (cos(x) + 1) = 0
2 cos(x) - 1 = 0 or cos(x) + 1 = 0
2 cos(x) = 1 or cos(x) = -1
cos(x) = 1/2 or cos(x) = -1
[x = arccos(1/2) + 2nπ or x = 2π - arccos(1/2) + 2nπ] or x = arccos(-1) + 2nπ
[x = π/3 + 2nπ or x = 5π/3 + 2nπ] or x = π + 2nπ
For 0 ≤ x ≤ 2π, the solutions are x = π/3, x = 5π/3, and x = π.
