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If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5},$ then
$(A)$ $\tan ^2 x=\frac{2}{3}$ $(B)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
$(C)$ $\tan ^2 x=\frac{1}{3}$ $(D)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$
$(A,C)$
$(A,B)$
$(B,C)$
$(D,B)$
Solution
$\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$
$\frac{\left(\sin ^2 x\right)^2}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5}$
$\text { Let } \sin ^2 x \text { bet, } t \geq 0$
$t \in[0,1]$
$\frac{ t ^2}{2}+\frac{(1-t)^2}{3}=\frac{1}{5}$
$\frac{ t ^2}{2}+\frac{1-2 t + t ^2}{3}=\frac{1}{5}$
$\frac{3 t ^2+2-4 t +2 t ^2}{6}=\frac{1}{5}$
$25 t ^2-20 t +10=6$
$25 t ^2-20 t +4=0$
$25 t ^2-10 t -10 t +4=0$
$(5 t -2)(5 t -2)=0$
$t =\frac{2}{5}$
$\therefore \sin ^2 x =\frac{2}{5}$
$\cos ^2 x =1-\sin ^2 x =1-\frac{2}{5}=\frac{3}{5}$
$\tan ^2 x =\frac{\sin ^2 x }{\cos ^2 x }=\frac{2}{3}$
