#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#
