[tex]\bf 2cos^2(x)+3cos(x)-2=0\impliedby \textit{so, notice is just a quadratic}
\\\\\\\
[2cos(x)~~-~~1][cos(x)~~+~~2]=0\\\\
-------------------------------\\\\
2cos(x)-1=0\implies 2cos(x)=1\implies cos(x)=\cfrac{1}{2}
\\\\\\
\measuredangle x=cos^{-1}\left( \frac{1}{2} \right)\implies \measuredangle x=
\begin{cases}
\frac{\pi }{3}\\\\
\frac{5\pi }{3}
\end{cases}\\\\
-------------------------------\\\\
cos(x)+2=0\implies cos(x)=-2[/tex]
now, for the second case, recall that the cosine is always a value between -1 and 1, so a -2 is just a way to say, such angle doesn't exist.
