$sin^2\left(x\right)-\sin^2\left(x\right)\cdot cos^2\left(x\right)=\tan^2\left(x\right)$
$3x\:+\:4y\:-\:x\:+\:2y\:$
$3y''+y=0$
$\frac{1}{1+2y}dy=\frac{-1}{4-x^2}dx$
$-168\::\:-7$
$x+5y=8$
$\frac{3}{4}\:=\:\frac{1}{4}\:x$
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