If frac{sin ^4 x}{2}+frac{cos ^4 x}{3}=frac{1 | vedclass

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Question

$\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$

$\frac{\left(\sin ^2 xight)^2}{2}+\frac{\left(1-\sin ^2 xight)^2}{3}=\frac{1}{5}$

$	e

Vedclass · Accepted Answer

$(A,B)$

Vedclass · Answer

$\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$

$\frac{\left(\sin ^2 xight)^2}{2}+\frac{\left(1-\sin ^2 xight)^2}{3}=\frac{1}{5}$

$	ext { Let } \sin ^2 x 	ext { bet, } t \geq 0$

$t \in[0,1]$

$\frac{ t ^2}{2}+\frac{(1-t)^2}{3}=\frac{1}{5}$

$\frac{ t ^2}{2}+\frac{1-2 t + t ^2}{3}=\frac{1}{5}$

$\frac{3 t ^2+2-4 t +2 t ^2}{6}=\frac{1}{5}$

$25 t ^2-20 t +10=6$

$25 t ^2-20 t +4=0$

$25 t ^2-10 t -10 t +4=0$

$(5 t -2)(5 t -2)=0$

$t =\frac{2}{5}$

$	herefore \sin ^2 x =\frac{2}{5}$

$\cos ^2 x =1-\sin ^2 x =1-\frac{2}{5}=\frac{3}{5}$

$	an ^2 x =\frac{\sin ^2 x }{\cos ^2 x }=\frac{2}{3}$

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

$(A)$ $\tan ^2 x=\frac{2}{3}$ $(B)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$

$(C)$ $\tan ^2 x=\frac{1}{3}$ $(D)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$

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If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5},$ then $(A)$ $\tan ^2 x=\frac{2}{3}$ $(B)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$ $(C)$ $\tan ^2 x=\frac{1}{3}$ $(D)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$