How do you solve #1+ cos x + cos 2x = 0#?

1 Answer
mason m
Apr 27, 2016

#x=pi/2+kpi,(2pi)/3+2kpi,(4pi)/3+2kpi,kinZZ#

Explanation:

Use the identity for the cosine double angle formula:

#cos2x=2cos^2x-1#

The equation then becomes

#1+cosx+2cos^2x-1=0#

#2cos^2x+cosx=0#

Factor out a #cosx# term.

#cosx(2cosx+1)=0#

Set each of these equal to #0#.

#cosx=0" "" ""and"" "" "2cosx+1=0#
#" "" "" "" "" "" "" "" "" "cosx=-1/2#

#cosx=0# occurs at #x=pi/2,(3pi)/2,(5pi)/2#, which can be generalized as #x=pi/2+kpi,kinZZ# (this means where #k# is an integer).

#cosx=-1/2# occurs at #x=(2pi)/3,(4pi)/3# and all other concentric angles (which are #2pi# away), thus we have #x=(2pi)/3+2kpi# and #x=(4pi)/3+2kpi#, both where #kinZZ#.

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