$\begin{aligned} & \text {} \because 3 \sin ^4 x+2 \cos ^4 x=\frac{6}{5} \\ & \Rightarrow 3 \sin ^4 x+2\left(1-\sin ^2 x\right)^2=\frac{6}{5} \\ & \Rightarrow 3 \sin ^4 x+2\left(1+\sin ^4 x-2 \sin ^2 x\right)=\frac{6}{5} \\ & \Rightarrow 25 \sin ^4 x-20 \sin ^2 x+4=0 \\ & \Rightarrow \sin ^2 x=\frac{20 \pm \sqrt{400-400}}{2 \times 25}=\frac{2}{5} \\ & \therefore \cos ^2 x=1-\frac{2}{5}=\frac{3}{5} \\ & \Rightarrow \sin 2 x=2 \sqrt{\frac{2}{5}} \sqrt{\frac{3}{5}}=\frac{2}{5} \sqrt{6} \\ & \& \cos 2 x=2 \times \frac{3}{5}-1=\frac{1}{5} \\ & \therefore \tan 2 x=2 \sqrt{6}\end{aligned}$