# Solve tan x + sec 2x = 1

Question
Solve $$\displaystyle{\tan{{x}}}+{\sec{{2}}}{x}={1}$$

2020-11-25
When $$\displaystyle{x}={0}:{\tan{{\left({0}\right)}}}+{\sec{{\left({2}\cdot{0}\right)}}}={1}$$ so x=0 is a solution, because $$\displaystyle{\tan{{\left({0}\right)}}}={0}{\quad\text{and}\quad}{\cos{{\left({0}\right)}}}{\quad\text{and}\quad}{\sec{{\left({0}\right)}}}={1}.$$
There is another solution at x=67.5 degrees. See below: If $$\displaystyle{y}={\sin{{\left({x}\right)}}}{t}{h}{e}{n}{w}{e}{c}{a}{n}{w}{\quad\text{or}\quad}{k}{o}{u}{t}{P}{S}{K}{\tan{{\left({x}\right)}}}.{P}{S}{K}{\cos{{\left({2}{x}\right)}}}={1}-{2}{{\sin}^{{2}}{\left({x}\right)}}={1}-{2}{y}^{{2}},{\sec{{\left({2}{x}\right)}}}=\frac{{1}}{{{1}-{2}{{\sin}^{{2}}{\left({x}\right)}}}},{\cos{{\left({x}\right)}}}=\sqrt{{{1}-{y}^{{2}}}}.{\sin{=}}{o}{p}\frac{{p}}{{h}}{y}{p}=\frac{{y}}{{1}}{s}{o}{a}{d}{j}=\sqrt{{{1}-{y}^{{2}}}}.{\tan{{\left({x}\right)}}}={o}{p}\frac{{p}}{{a}}{d}{j}=\frac{{y}}{\sqrt{{{1}-{y}^{{2}}}}}$$.
Therefore, $$\displaystyle{\tan{{\left({x}\right)}}}+{\sec{{\left({2}{x}\right)}}}=\frac{{y}}{\sqrt{{{1}-{y}^{{2}}}}}+\frac{{1}}{{{1}-{2}{y}^{{2}}}}={1}.$$
$$\displaystyle\frac{{y}}{\sqrt{{{1}-{y}^{{2}}}}}={1}-\frac{{1}}{{{1}-{2}{y}^{{2}}}}=\frac{{{1}-{2}{y}^{{2}}-{1}}}{{{1}-{2}{y}^{{2}}}}=-{2}\frac{{y}^{{2}}}{{{1}-{2}{y}^{{2}}}}.$$
y=0 is a solution $$\displaystyle{\left({\sin{{\left({x}\right)}}}={0}{s}{o}{x}={0}\right)}$$and:
$$\displaystyle\frac{{1}}{\sqrt{{{1}-{y}^{{2}}}}}=-{2}\frac{{y}}{{{1}-{2}{y}^{{2}}}},$$
$$\displaystyle\sqrt{{{1}-{y}^{{2}}}}=\frac{{{2}{y}^{{2}}-{1}}}{{2}}{y}.$$
Squaring: $$\displaystyle{1}-{y}^{{2}}=\frac{{{4}{y}^{{4}}-{4}{y}^{{2}}+{1}}}{{4}}{y}^{{2}},{4}{y}^{{2}}-{4}{y}^{{4}}={4}{y}^{{4}}-{4}{y}^{{2}}+{1},{8}{y}^{{4}}-{8}{y}^{{2}}+{1}={0}.$$
So $$\displaystyle{y}^{{2}}={\left({8}\pm\frac{\sqrt{{{64}-{32}}}}{{16}}=\frac{{1}}{{2}}\pm\frac{\sqrt{{2}}}{{4}}={0.8536}{\quad\text{or}\quad}{0.1464}\right.}$$ approx and y=0.9239 or $$\displaystyle{0.3827}={\sin{{\left({x}\right)}}}.$$
Therefore $$\displaystyle{x}=\pm{67.5}º,{x}=\pm{22.5}º$$, but we may have to eliminate "solutions" that don't fit the original equation.
These are the solutions that do fit: 67.5, -22.5 and of course 0. The periodic nature of the trig functions gives us a series: 0, 67.5, -22.5, 157.5, 180, 247.5, 337.5, etc.

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