Help for a basic calculus problem

If f(x)= cot (2x), what are all values of x in which f(x)=f '' (x)?
Domain of the function is larger or equal to 0 and smaller or equal to 2pi

f(x) = cot(2x)

f ' (x) = - 2csc^2(2x)

f '' (x) = -2[2csc(2x)(-csc(2x) cot(2x)*2) ]

= 8csc^2(2x)cot(2x)

=> cot(2x) = 8 csc^2(2x)cot(2x)

=> cot(2x) [1 - 8csc^2(2x) ] = 0

case 1: cot(2x) = 0

2x = π/2, 3π/2, 5π/2, 7π/2

x = π/4, 3π/4, 5π/4, 7π/4

case 2: 8csc^2(2x) = 1
csc^2(2x) = 1/8

csc(2x) = (1/4)√2

sin(2x) = 2√2

there is no solution sin x value lies between - 1 and 1

So x = π/4, 3π/4, 5π/4 and 7π/4