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