$10\cos\left(2x\right)-5=0$
$\frac{1-2sin\left(x\right)+sin^2\left(x\right)}{1-sin^2\left(x\right)}=\frac{1-sin\left(x\right)}{1+sin\left(x\right)}$
$x^2-10x+25=1$
$\left(2a-8\right)-\left(5a-7\right)$
$x+2+-4x^2+16x+6$
$\frac{-u+1}{-u-1}$
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