Solve the equation. Give exact solutions which lie in [0, 2π).?

csc³(x) + csc²(x) = 4 csc(x) + 4
csc³(x) + csc²(x) - 4 csc(x) - 4 = 0
css²(x) [csc(x) + 1] - 4 [csc(x) + 1] = 0
[csc(x) + 1] [csc²(x) - 4] = 0
[csc(x) + 1] [csc(x) + 2] [csc(x) - 2] = 0
csc(x) = -1 or csc(x) = -2 or csc(x) = 2
1/sin(x) = -1 or 1/sin(x) = -2 or 1/sin(x) = 2
sin(x) = -1 or sin(x) = -1/2 or sin(x) = 1/2
x = 3π/2 or x = π + (π/6), 2π - (π/6) or x = π/6, π - (π/6)
x = π/6, 5π/6, 7π/6, 3π/2, 11π/6

csc^3(x)+csc^2(x)=4csc(x)+4.(1). Put (s,c)=[sin(x),cos(x)]. Multiply (1) by s^3
getting 1 + s = 4s^2+4s^3, ie., 4s^2(s+1) = 1(s+1), ie., (s+1)(1+2s)(1-2s) = 0.
Then s = -1...(i) or s = -(1/2)...(ii) or s = +(1/2)...(iii).  x is in [0,2pi).

For (i)   holding, x = (3/2)pi.
For (ii)  holding, x = (7/6)pi or (11/6)pi.
For (iii) holding, x = (1/6)pi or   (5/6)pi.