#cos(2x) = 2cos^2(x)-1##color(white)("XXXX")#(This is one version of the double angle formula for cos)
So
#color(white)("XXXX")##cos(2x)+cos(x)-2 =0#
can be written as
#color(white)("XXXX")##2cos^2(x)+cos(x)-3 = 0#
This can be factored as:
#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#
But #-3/2# is not within the range of #cos(x)# for any value of #(x)#.
#color(white)("XXXX")#(#cos(x) in [-1,+1]#)
So #cos(x) = 1#
#color(white)("XXXX")#within the range #x in [0, 2pi)#
#x=0#
or, in general
#x=2npi# for #AA n in ZZ#
