How do you find the general solutions for #cos(2x) + 5cos(x) + 3 = 0#?

1 Answer
Aug 18, 2015

For #x in [0,2pi]#
#color(white)("XXXX")##x=pi#

Explanation:

#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#