Given:
[tex]6\cos^2\theta+\cos\theta-2=0[/tex]Let
[tex]\cos\theta=t[/tex]Then
[tex]\begin{gathered} 6t^2+t-2=0 \\ (2t-1)(3t+2)=0 \\ 2t-1=0 \\ \Rightarrow t=\frac{1}{2} \\ \\ 3t+2=0 \\ \Rightarrow t=-\frac{2}{3} \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} \cos\theta=\frac{1}{2} \\ \\ \Rightarrow\theta=60^o \\ \\ \cos\theta=-\frac{2}{3} \\ \Rightarrow\theta=131.8^o \end{gathered}[/tex]