$\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$

$\frac{\left(\sin ^2 x\right)^2}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5}$

$\text { Let } \sin ^2 x \text { bet, } t \geq 0$

$t \in[0,1]$

$\frac{ t ^2}{2}+\frac{(1-t)^2}{3}=\frac{1}{5}$

$\frac{ t ^2}{2}+\frac{1-2 t + t ^2}{3}=\frac{1}{5}$

$\frac{3 t ^2+2-4 t +2 t ^2}{6}=\frac{1}{5}$

$25 t ^2-20 t +10=6$

$25 t ^2-20 t +4=0$

$25 t ^2-10 t -10 t +4=0$

$(5 t -2)(5 t -2)=0$

$t =\frac{2}{5}$

$\therefore \sin ^2 x =\frac{2}{5}$

$\cos ^2 x =1-\sin ^2 x =1-\frac{2}{5}=\frac{3}{5}$

$\tan ^2 x =\frac{\sin ^2 x }{\cos ^2 x }=\frac{2}{3}$