`tan(x+pi/6) = 2 tanx`
`=> (tanx+1/sqrt3)/(1-tanx*1/sqrt3) = 2 tanx`
`=>(sqrt3tanx+1)/(sqrt3 - tanx) = 2tanx`
`=>2tan^2x - sqrt3tanx+1 =0`
This is a quadratic equation.
So, its discriminant , `D = (sqrt3)^2 - 4(1)(2) = 3-8 = -5`
As `D lt 0`, no real roots exists for the given equation.
