Inequalities find the set of values of x

[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.

 

[SammyS]

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.

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 ?

 

[synkk]

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

 

[SammyS]

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

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

 

[synkk]

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

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

is that the correct answer then?

 

[SammyS]

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

is that the correct answer then?

Looks good !

 

[synkk]

Looks good !

Thanks!