Given
[tex]2\sec ^2(x)-13\tan (x)=-13[/tex]Add 13 to both sides
[tex]\begin{gathered} 2\sec ^2(x)-13\tan (x)+13=-13+13 \\ 2\sec ^2(x)-13\tan (x)+13=0 \end{gathered}[/tex]We have that
[tex]\sec ^2(x)=1+\tan ^2(x)[/tex]So, substitute in the above equation
[tex]2(1+\tan ^2(x))-13\tan (x)+13=0[/tex]Simplify
[tex]\begin{gathered} 2+2\tan ^2(x)-13\tan (x)+13=0 \\ 15+2\tan ^2(x)-13\tan (x)=0 \end{gathered}[/tex]Reordering the equation
[tex]2\tan ^2(x)-13\tan (x)+15=0[/tex]We get a quadratic equation, then solve by factoring
[tex](2\tan (x)-3)(\tan (x)-5)=0[/tex]Separate the solutions
[tex]\begin{gathered} 2\tan (x)-3=0 \\ 2\tan (x)-3+3=0+3 \\ 2\tan (x)=3 \\ \frac{2\tan (x)}{2}=\frac{3}{2} \\ \tan (x)=\frac{3}{2} \end{gathered}[/tex]And
[tex]\begin{gathered} \tan (x)-5=0 \\ \tan (x)-5+5=0+5 \\ \tan (x)=5 \end{gathered}[/tex]Next, solve for x for each solution
[tex]\begin{gathered} \tan (x)=\frac{3}{2} \\ x=\tan ^{-1}(\frac{3}{2}) \\ x=56.3 \end{gathered}[/tex]And
[tex]\begin{gathered} \tan (x)=5 \\ x=\tan ^{-1}(5) \\ x=78.7 \end{gathered}[/tex]Answer:
x = 56.3° and x = 78.7°