How do you solve #cos (x/2 )- cosx=0# if #0°<=x<360#?

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Answer 1 · 2016-10-07T17:23:39

Solution is #{0^o,240^o}#

As #cosx=2cos^2(x/2)-1#

#cos(x/2)-cosx=0# can be written as

#cos(x/2)-(2cos^2(x/2)-1)=0#

or #-2cos^2(x/2)+cos(x/2)+1=0#

or #2cos^2(x/2)-cos(x/2)-1=0#

or #2cos^2(x/2)-2cos(x/2)+cos(x/2)-1=0#

or #2cos(x/2)(cos(x/2)-1)+1(cos(x/2)-1)=0#

or #(2cos(x/2)+1)(cos(x/2)-1)=0#

Therefore, either #2cos(x/2)+1=0# i.e. #cos(x/2)=-1/2# and #x/2=2npi+-(2pi)/3# or #x=4npi+-(4pi)/3#, where #n# is an integer.

or #cos(x/2)-1=0# i.e #cos(x/2)=1# and #x/2=2npi# i.e. #x=4npi# where #n# is an integer

In the range #0^o<=x<=360^0#, the solution is #{0^o,240^o}#

Answer 2 · 2016-12-31T03:47:10

#x={0^@,240^@}#

#cos(x/2)-cosx=0#

#=>cosx=cos(x/2)#

#=>x=2npipm(x/2)" where "n in ZZ#

when

#=>x=2npi+(x/2)#

#=>x/2=2npi#

#=>x=4npi#

as #0<\x<=360^@ #for n=0; #x=0^@#

when

#=>x=2npi-(x/2)#

#=>x+x/2=2npi#

#=>(3x)/2=2npi#

#=>x=(4npi)/3#

as #0<\x<=360^@ #for n=1; #x=240^@#