# Solve tan x + sec 2x = 1

Solve $\mathrm{tan}x+\mathrm{sec}2x=1$
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hajavaF

When $x=0:\mathrm{tan}\left(0\right)+\mathrm{sec}\left(2\cdot 0\right)=1$ so x=0 is a solution, because $\mathrm{tan}\left(0\right)=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}\left(0\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{sec}\left(0\right)=1.$
There is another solution at x=67.5 degrees. See below: If $y=\mathrm{sin}\left(x\right)\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mathrm{tan}\left(x\right).\mathrm{cos}\left(2x\right)=1-2{\mathrm{sin}}^{2}\left(x\right)=1-2{y}^{2},\mathrm{sec}\left(2x\right)=\frac{1}{1-2{\mathrm{sin}}^{2}\left(x\right)},\mathrm{cos}\left(x\right)=\sqrt{1-{y}^{2}}.\mathrm{sin}=op\frac{p}{h}yp=\frac{y}{1}soadj=\sqrt{1-{y}^{2}}.\mathrm{tan}\left(x\right)=op\frac{p}{a}dj=\frac{y}{\sqrt{1-{y}^{2}}}$.
Therefore, $\mathrm{tan}\left(x\right)+\mathrm{sec}\left(2x\right)=\frac{y}{\sqrt{1-{y}^{2}}}+\frac{1}{1-2{y}^{2}}=1.$
$\frac{y}{\sqrt{1-{y}^{2}}}=1-\frac{1}{1-2{y}^{2}}=\frac{1-2{y}^{2}-1}{1-2{y}^{2}}=-2\frac{{y}^{2}}{1-2{y}^{2}}.$
y=0 is a solution $\left(\mathrm{sin}\left(x\right)=0sox=0\right)$and:
$\frac{1}{\sqrt{1-{y}^{2}}}=-2\frac{y}{1-2{y}^{2}},$
$\sqrt{1-{y}^{2}}=\frac{2{y}^{2}-1}{2}y.$
Squaring: $1-{y}^{2}=\frac{4{y}^{4}-4{y}^{2}+1}{4}{y}^{2},4{y}^{2}-4{y}^{4}=4{y}^{4}-4{y}^{2}+1,8{y}^{4}-8{y}^{2}+1=0.$
So ${y}^{2}=\left(8±\frac{\sqrt{64-32}}{16}=\frac{1}{2}±\frac{\sqrt{2}}{4}$