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

Question

66546 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]}

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