$.91^5$
$2\sin^2\:x-1=0$
$\frac{1x^3+1x^2+1x^1+1}{x^2-2}$
$\lim_{x\to3}\left(2x^2-6x+1\right)$
$1+cos\left(2x\right)=\tan\left(x\right)sin\left(2x\right)$
$m^2=0$
$\frac{\tan\left(-x\right)}{\sec\left(x\right)}$
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