Question

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

Answer 1

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`