Find all real numbers in the interval #[0, 2pi)# round to the nearest tenth? #3 sin^2x=sin x#

Question

1047 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-17T12:07:15

#x=0^c,0.34^c,pi^c,2.80^c#

Rearrange to get:
#3sin^2x-sinx=0#

#sinx=(1+-sqrt(1^2))/6#

#sinx=(1+1)/6 or (1-1)/6#

#sinx=2/6 or 0/6#

#sinx=1/3or0#

#x=sin^-1(0)=0, pi-0=0^c,pi^c#
or
#x=sin^-1(1/3)=0.34, pi-0.34=0.34^c,2.80^c#

#x=0^c,0.34^c,pi^c,2.80^c#