4cos²x=ctgx
ОДЗ
sinx=0
x=πk k∈Z

4cos²x=cosx/sinx
4cos²x-cosx/sinx=0
cosx(4cosx-1/sinx*)=0
cosx(4cosx*sinx-1)/sinx=0
cosx(2sin2x-1)/sinx=0
2cosx(sin2x-0.5)/sinx=0
cosx=π/2+πk, k∈Z
sin2x=1/2
2x=π/3+2πk, k∈Z
x=π/6+πk      k∈Z
2x=2π/3+2πk, k∈Z
x=π/3+πk  k∈Z

Ответ
x=π/3+πk  k∈Z
x=π/6+πk      k∈Z
cosx=π/2+πk, k∈Z