How would you solve tan2xcotx-3=0? Thanks in advance :)

Tan2x.cotgx - 3 = 0
<=> sin2x.cosx / cos2x.sinx - 3 = 0
We know :
{ cos2x # 0 => 2x # π/2 + kπ => x # π/4 + kπ/2 ( k of Z )
{ sinx # 0 => x # kπ

<=> sin2x.cosx - 3cos2x.sinx = 0
<=> 2sinxcos²x - 3cos2x.sinx = 0
<=> sinx( 2cos²x - 3cos2x ) = 0

<=> sinx = 0 ( Don't choose because sinx # 0 )
<=> 2cos²x - 3cos2x = 0
<=> 2cos²x - 3(2cos²x - 1) = 0
<=> 2cos²x - 6cos²x + 3 = 0
<=> -4cos²x = -3
<=> cos²x = 3/4

<=> cosx = √3/2
<=> cosx = -√3/2

<=> x = ±π/6 + k2π ( k of Z )
<=> x = ±5π/6 + k2π