How do you solve #2cos^2x+cosx=0# and find all exact general solutions?

Question

52572 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-16T07:04:33

General Solution for #2cos^2x+cosx=0# is

#x=((2n+1)pi)/2# or #x=2npi+-(2pi)/3#, where #n# is an integer.

#2cos^2x+cosx=0#

#hArrcosx(2cosx+1)=0#

i.e. either #cosx=0# or #2cosx+1=0# i.e. #cosx=-1/2#

General solution for #cosx=0# is #x=((2n+1)pi)/2#, where #n# is an integer

and general solution for #cosx=-1/2=cos(+-(2pi)/3)# is #x=2npi+-(2pi)/3#, where #n# is an integer

Hence, General Solution for #2cos^2x+cosx=0# is

#x=((2n+1)pi)/2# or #x=2npi+-(2pi)/3#, where #n# is an integer.

Answer 2 · 2016-12-16T07:07:08

The solutions are #S={pi/2+2pin,3/2pi+2pin,2/3pi+2pin,4/3pi+2pin} #

The equation is

#2cos^2+cosx=0#

#cosx(2cosx+1)=0#

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

#cosx=0# gives #x=pi/2+2pin# and #x=3/2pi+2pin#

#cosx=-1/2# gives #x=2/3pi+2pin# and #x=4/3pi+2pin#