How do you find all solutions for #sin 2x = cos x# for the interval #[0,2pi]#?

Question

62724 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-28T13:52:34

The solutions are #S={1/2pi, 3/2pi, 1/6pi, 5/6pi}#

We need

#sin2x=2sinxcosx#

Therefore,

#sin2x=cosx#

#sin2x-cosx=0#

#2sinxcosx-cosx=0#

#cosx(2sinx-1)=0#

So,

#{(cosx=0),(2sinx-1=0):}#

#<=>#, #{(cosx=0),(sinx=1/2):}#

#<=>#, #{(x=pi/2 , 3/2pi),(x=1/6pi, 5/6pi):}# #AA x in [0, 2pi]#

The solutions are #S={1/2pi, 3/2pi, 1/6pi, 5/6pi}#

graph{sin(2x)-cosx [-1.622, 9.475, -2.51, 3.04]}