How do you solve #2cos^2x-cosx-1=0# and find all solutions in the interval #[0,2pi)#?

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Answer 1 · 2017-01-15T07:23:52

Solution is #{0,(2pi)/3,(4pi)/3}#

#2cos^2x-cosx-1=0# can be written as

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

or #2cosx(cosx-1)+1(cosx-1)=0#

or #(2cosx+1)(cosx-1)=0#

#:.# either #2cosx+1=0# i.e. #cosx=-1/2# and in the interval #[0,2pi)# #x=(2pi)/3# or #(4pi)/3#

or #cosx-1=# i.e. #cosx=1# and in given interval #x=0#.

Hence solution is #{0,(2pi)/3,(4pi)/3}#