∵tanx+1tanx=4
∴tan²x-4tanx+1=0
∴tanx=[4±√(16-4)]/2=2±√3
∴sin2x=2tanx/(1+tan²x)
=2(2±√3)/[1+(2±√3)²]
=(4±2√3)/(8±2√3)
=(2±√3)/(4±√3)
∵tanx+1tanx=4
∴tan²x-4tanx+1=0
∴tanx=[4±√(16-4)]/2=2±√3
∴sin2x=2tanx/(1+tan²x)
=2(2±√3)/[1+(2±√3)²]
=(4±2√3)/(8±2√3)
=(2±√3)/(4±√3)