Solve for: tan(θ)^2+sec(θ)=1

Expression: ${\tan\left({θ}\right)}^{2}+\sec\left({θ}\right)=1$

Determine the defined range

$\begin{array} { l }{\tan\left({θ}\right)}^{2}+\sec\left({θ}\right)=1,& \begin{array} { l }θ≠\frac{ π }{ 2 }+kπ,& k \in ℤ\end{array}\end{array}$

Use ${\tan\left({t}\right)}^{2}={\sec\left({t}\right)}^{2}-1$ to expand the expression

${\sec\left({θ}\right)}^{2}-1+\sec\left({θ}\right)=1$

Move the constant to the left-hand side and change its sign

${\sec\left({θ}\right)}^{2}-1+\sec\left({θ}\right)-1=0$

Calculate the difference

${\sec\left({θ}\right)}^{2}-2+\sec\left({θ}\right)=0$

To get an equation that is easier to solve, substitute $t$ for $\sec\left({θ}\right)$

${t}^{2}-2+t=0$

Solve the equation for $t$

$\begin{array} { l }t=-2,\\t=1\end{array}$

Substitute back $t=\sec\left({θ}\right)$

$\begin{array} { l }\sec\left({θ}\right)=-2,\\\sec\left({θ}\right)=1\end{array}$

Solve the equation for $θ$

$\begin{array} { l }\begin{array} { l }θ=\frac{ 2π }{ 3 }+2kπ,& k \in ℤ\end{array},\\\begin{array} { l }θ=\frac{ 4π }{ 3 }+2kπ,& k \in ℤ\end{array},\\\sec\left({θ}\right)=1\end{array}$

Solve the equation for $θ$

$\begin{array} { l }\begin{array} { l }θ=\frac{ 2π }{ 3 }+2kπ,& k \in ℤ\end{array},\\\begin{array} { l }θ=\frac{ 4π }{ 3 }+2kπ,& k \in ℤ\end{array},\\\begin{array} { l }θ=2kπ,& k \in ℤ\end{array}\end{array}$

Find the union

$\begin{array} { l }\begin{array} { l }θ=\frac{ 2kπ }{ 3 },& k \in ℤ\end{array},& \begin{array} { l }θ≠\frac{ π }{ 2 }+kπ,& k \in ℤ\end{array}\end{array}$

Find the intersection of the solution and the defined range

$\begin{array} { l }θ=\frac{ 2kπ }{ 3 },& k \in ℤ\end{array}$