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

Question

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

1 Answer

Impact of this question

3181 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-07T15:28:19

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 $[pi/6+2kpi, 5/6pi+2kpi]uu[pi+2kpi, 2pi +2kpi]$

$kin ZZ$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2sin^2x>=sinx2sin^2x>=sinx using a sign chart?

1 Answer

Explanation:

Related questions

Impact of this question

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