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