Solve tan x + sec 2x = 1

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

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