# Solve: cos 2 </msup> &#x2061;<!-- ⁡ --> <mrow class="MJX-TeXAtom-ORD">

Solve: ${\mathrm{cos}}^{2}x>{\mathrm{sin}}^{2}x$
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Xzavier Shelton
${\mathrm{cos}}^{2}x>{\mathrm{sin}}^{2}x\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}2x={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x>0$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}2n\pi -\frac{\pi }{2}<2x<2n\pi +\frac{\pi }{2}$
where n is any integer
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Hector Petersen
my approach is to divide both sides by ${\mathrm{cos}}^{2}\left(x\right)$ to get
${\mathrm{tan}}^{2}\left(x\right)<1$
or equivalently
$-1<\mathrm{tan}\left(x\right)<1$
Remember that $\mathrm{tan}\left(x+\pi \right)=\mathrm{tan}\left(x\right)$