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How do you solve for x in #cos2x+2cosx+1=0#?

1 Answer

Feb 13, 2015

#cos(2x)+2cos(x)+1 = 0# when
# x = pi/2# (plus #n pi# for all integer #n#)
or
#x = pi# (plus #2n pi# for all integer #n#)

How we got this :
#cos(2x) + 2cos(x) + 1 = 0#
#2cos^2(x) - 1 + 2cos(x )+ 1 = 0# by a double angle formula for #cos(2x)#

#(2) (cos^2(x) + cos(x) ) = 0#
#(2) (cos(x)) (cos(x)+1) = 0#

So
#cos(x) = 0# which gives #x = pi/2, (3pi)/2, (5pi)/2# etc.
or
#cos(x) = -1# which gives #x = pi, 3pi, 5pi# etc.

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