Squaring: 1+sin2x=41⟹sin2x=−43, and since 0<x<π, 2π<x≤π (as cosx+sinx=21>0 narrows this to 2π<x<π).
\[\frac{2\tan x}{1+\tan^2x} = -\frac34 \implies 3\tan^2x+8\tan x+3=0 \implies \tan x = \frac{-8\pm\sqrt{64-36}}6 = \frac{-4\pm\sqrt{7}}3\]
Since 2π<x<π, tanx<0, so:
\[\tan x = -\frac{4+\sqrt{7}}3\]