# The general solution of tan(x)=tan(2x−pi/2)

The general solution of $\mathrm{tan}\left(x\right)=\mathrm{tan}\left(2x-\pi /2\right)$
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Savanna Smith
$\mathrm{tan}\left(2x-\frac{\pi }{2}\right)=-\mathrm{tan}\left(\frac{\pi }{2}-2x\right)=-\frac{1}{\mathrm{tan}2x}$
$\mathrm{tan}x=-\frac{1}{\mathrm{tan}2x}\to \mathrm{tan}x\mathrm{tan}2x=-1$
$\mathrm{tan}x\frac{2\mathrm{tan}x}{1-{\mathrm{tan}}^{2}x}=-1\to 2{\mathrm{tan}}^{2}x=-1+{\mathrm{tan}}^{2}x\to {\mathrm{tan}}^{2}x=-1!$
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Alberto Calhoun
If you set $t=\mathrm{tan}x$ and apply the duplication formula for the tangent, you have
$\mathrm{tan}\left(2x-\frac{\pi }{2}\right)=-\frac{1}{\mathrm{tan}2x}=\frac{{t}^{2}-1}{2t},$
so the equation becomes
$\frac{{t}^{2}-1}{2t}=t\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{t}^{2}-1=2{t}^{2}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{t}^{2}=-1,$
which has no real solution