# Inequalities find the set of values of x

1. Oct 25, 2012

### synkk

find the set of values of x for which $$2x > \dfrac{3x + 1}{x+1}$$ done this and got $$-1<x<-0.5, x > 1$$

b) find the set of inequalities where

$$2sint > \dfrac{3sint + 1}{sint + 1}$$ where $$-\pi < t < \pi$$

first I found the set values of t suitable in the range for -1,-0.5,1 which I got to be as $$-\frac{\pi}{2}, - \frac{\pi}{6}, - \frac{5\pi}{6}, \frac{pi}{2}$$ and hence getting $$\frac{-5\pi}{6} < t < \frac{-\pi}{6}, t > \frac{\pi}{2}$$

however I'm not sure if it is correct, and if it isn't I don't know how else to do it.

Last edited: Oct 25, 2012
2. Oct 25, 2012

### SammyS

Staff Emeritus
You did the first part correctly; the part where you solved for x.

You can directly substitute sin(t) for x in that result.

$\displaystyle -1<\sin(t)<-0.5\,,\ \sin(t)>1$

I sin(t) ever greater than 1 ?

3. Oct 25, 2012

### synkk

No it's not, also are the inequalities supposed to be in terms of t?

4. Oct 25, 2012

### SammyS

Staff Emeritus
Yes. Your final answer should be in terms of t, but you can eliminate sin(t) > 1, from contributing anything to the solution.

5. Oct 25, 2012

### synkk

$$\frac{-5\pi}{6} < t < \frac{-\pi}{6}$$

is that the correct answer then?

6. Oct 25, 2012

### SammyS

Staff Emeritus
Looks good !

7. Oct 25, 2012

Thanks!