What are the solutions to 2cos^2x=cosx on the interval [0,2pi)?

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Answer 1 · 2017-12-07T15:49:15

The solutions are #S={1/3pi, 1/2pi, 3/2pi, 5/3pi}#

We need

The equation is

#2cos^2x=cosx#

#2cos^2x-cosx=0#

Factorising

#cosx(2cosx-1)=0#

#cosx=0#, #=>#, #x in{pi/2 , 3/2pi}#, #mod[2pi]#

#2cosx-1=0#, #=>#, #cosx=1/2#, #=>#, #x in {pi/3, 5/3pi}#, #[mod 2pi]#