[tex]2cos^2x+5cosx+2=0\\
x\in<-1;1>\\
\Delta=b^2-4ac\\
\Delta=9\\
\sqrt{\Delta}=3\\
cosx=1\\
or\\
cosx=-6\\
-6\notin<-1;1>\\[/tex]
Then using trygonometric table or a graph we read that:
[tex]cosx=1\Leftrightarrow(x)=0[/tex]
Period of cosinus function = 2π
So the last answer is [tex]x=0+2k\pi\\[/tex]
