Please help me with trigonometry

Solve the equation -cotx = 4cos^2 x
for values of x are in between 0 and pi

I've gotten only 1 of the 3 answers which is pi/2. :(

cot(x) = cos(x) / sin(x)

- 1 / sin(x) = 4 cos(x)

-1/4 = sin(x) cos(x)

-1/4 = 1/2 sin(2x)

-1/2 = sin(2x)

2x = (-1/6 π) + (π) = 5/6 π
2x = (-1/6 π) + (2π) = 11/6π

x = 5/12 π
x = 11/12 π

these are the other 2 answers

-ctg(x) = 4cos²(x)
-cos(x)/sin(x) = 4cos²(x)
sin(x)≠0 => x≠πk

-cos(x)/sin(x) = 4cos²(x)
4cos²(x)+cos(x)/sin(x) = 0
cos(x)[4cos(x)+1/sin(x)] = 0
cos(x)=0 => x₁=π/2+πk
and
4cos(x)+1/sin(x)=0
4cos(x)sin(x)+1=0
2sin(2x)=-1
sin(2x)=-1/2 => 2x=-π/6+2πk => x₂=-π/12+πk
2x=π+π/6+2πk => x₃=7π/12+πk


x₁=π/2+πk ∩ [0,π] = {π/2}
x₂=-π/12+πk ∩ [0,π] = {11π/12}
x₃=7π/12+πk ∩ [0,π] = {7π/12}

pi/2