#cos(2x) = 2cos^2(x)-1##color(white)("XXXX")#(one of the double angle formulae)

#cos(2x)+5cos(x)+3=0#

#rArr##color(white)("XXXX")##2cos^2(x)+5cos(x)+3=0#

Factoring:

#rArr##color(white)("XXXX")##(2cos(x)+3)(cos(x)+1)=0#

So

#cos(x) = -3/2##color(white)("XXXX")#or#color(white)("XXXX")##cos(x)=-1#

Since #cos(x) in [-1,+1]# for all values of #x#

#color(white)("XXXX")##cos(x)=-3/2# is an extraneous result.

#cos(x) = -1#

#rArr##color(white)("XXXX")##x= 3pi# within the range #[0,2pi]#

if the range is not restricted:

#color(white)("XXXX")##x = pi+2npi, AAn in ZZ#

Note:

I modified a term in the question from (#cos2(x)#) to (#cos(2x)#);

the other possible interpretation might have been (#cos^2(x)#)

but that version has no solutions for #x#