How do you solve #2sin^2x>=sinx# using a sign chart?

1 Answer
Mar 7, 2017

Answer:

The solution is #[pi/6+2kpi, 5/6pi+2kpi]uu[pi+2kpi, 2pi +2kpi]#
#k in ZZ#

Explanation:

Let's rearrange and factorise the inequality

#2sin^2x>=sinx#

#2sin^2x-sinx>=0#

#sinx(2sinx-1)>=0#

Let #f(x)=sinx(2sinx-1)#

We need to find the important points for the sign chart

#sinx=0#, #=>#, #x=0,pi,2pi#

#sinx=1/2#, #=>#, #x=pi/6, 5pi/6#

Now, we can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaaaaaa)##0##color(white)(aaaa)##pi/6##color(white)(aaaa)##5/6pi##color(white)(aaaaaaaa)##pi##color(white)(aaaaaa)##2pi#

#color(white)(aaaa)##sinx##color(white)(aaaaaaaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)####color(white)(aaa)##+##color(white)(aaaaa)##-#

#color(white)(aaaa)##2sinx-1##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)####color(white)(aaa)##-##color(white)(aaaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)####color(white)(aaa)##-##color(white)(aaaaa)##+#

Therefore,

#f(x)>=0# when #x in [pi/6, 5/6pi]uu [pi,2pi]#

The solution is #[pi/6+2kpi, 5/6pi+2kpi]uu[pi+2kpi, 2pi +2kpi]#

#kin ZZ#