Question

Solve 2cos^2x+5cosx+2=0

Answer 1

`2cos^2x+5cosx+2=0\\ x\in<-1;1>\\ \Delta=b^2-4ac\\ \Delta=9\\ √(\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\\`