Thread: What values of x satisy the inequality

1. What values of x satisy the inequality

Sin2x<Sinx so that x is between 0 and 2Pi

So
Sin2x-sinx=0
2sinx.cosx-sinx
sinx(2cosx-1)=0
Sinx=0 or cosx=1/2

x=0,Pi/3,Pi, 5Pi/3 and 2Pi

So how do I answer this question from here?
Should I just substitute a value between 0 and Pi/3, Pi/3 and Pi, Pi and 5Pi/3....... and see whether sin2x value is smaller than the sinx value?

Is this a good way to do it or can you suggest a more expedient method to answer this Q? My method is rather time consuming I think - wondering if there's a better methd

2. Originally Posted by Utterconfusion
Sin2x<Sinx so that x is between 0 and 2Pi

So
Sin2x-sinx=0
2sinx.cosx-sinx
sinx(2cosx-1)=0
Sinx=0 or cosx=1/2

x=0,Pi/3,Pi, 5Pi/3 and 2Pi

So how do I answer this question from here?
Should I just substitute a value between 0 and Pi/3, Pi/3 and Pi, Pi and 5Pi/3....... and see whether sin2x value is smaller than the sinx value?

Is this a good way to do it or can you suggest a more expedient method to answer this Q? My method is rather time consuming I think - wondering if there's a better method
Your method, finding the zeros of the function $\displaystyle f(x) = \sin(2x) - \sin x$, is as good as any. Notice that these are all simple zeros (no repeated roots), so the graph of f(x) will cross the axis at each of these zeros. Therefore it will be alternately positive and negative in the intervals (0,π/3), (π/3,π), (π,5π/3) and (5π/3,2π). All you need to do is to check that f(x) is positive in the first of those intervals.

3. Originally Posted by Utterconfusion
Sin2x<Sinx so that x is between 0 and 2Pi

So
Sin2x-sinx=0
2sinx.cosx-sinx
sinx(2cosx-1)=0
Sinx=0 or cosx=1/2

x=0,Pi/3,Pi, 5Pi/3 and 2Pi

So how do I answer this question from here?
Should I just substitute a value between 0 and Pi/3, Pi/3 and Pi, Pi and 5Pi/3....... and see whether sin2x value is smaller than the sinx value?

Is this a good way to do it or can you suggest a more expedient method to answer this Q? My method is rather time consuming I think - wondering if there's a better methd
HI

Those 5 solutions of x which you found would divide the number line into a few sub intervals . Just substitute a value and check whether its validity .

Example , for (0 , pi/3)

just pick a value in between , and substitute into sinx(2cosx-1) and if its < 0 , then its valid .

4. Originally Posted by Opalg
Your method, finding the zeros of the function $\displaystyle f(x) = \sin(2x) - \sin x$, is as good as any. Notice that these are all simple zeros (no repeated roots), so the graph of f(x) will cross the axis at each of these zeros. Therefore it will be alternately positive and negative in the intervals (0,π/3), (π/3,π), (π,5π/3) and (5π/3,2π). All you need to do is to check that f(x) is positive in the first of those intervals.
Hi. sorry I'm not that good at Math so I'm a bit confused over something. I guess sometimes I tend to get all mechanical and just do something to obtain an answer w/out thinking why I'm doing it; I guess this is one of those times.

Now What I was thinking was of the 2 graphs Sin2x and sinx both graphs drawn on the same (set of axis?)......or both drawn together, and where the sin2x dips below the sinx--->Answer to Question. But when we combine the 2 or whatever to F(x)=Sin2x-Sinx, I can't seem to relate the 2. What i mean is I can't figure out how just because when f(x)<0 it means that the Sin2x graph dips below the sinx graph.

Now I'm really confused. Now I'm wondering why at all I took Sin2x<Sinx as
Sin2x-sinx=0 and found the roots of it!

Can you explain to me whats happening here please?

5. Originally Posted by Opalg
Your method, finding the zeros of the function $\displaystyle f(x) = \sin(2x) - \sin x$, is as good as any. Notice that these are all simple zeros (no repeated roots), so the graph of f(x) will cross the axis at each of these zeros. Therefore it will be alternately positive and negative in the intervals (0,π/3), (π/3,π), (π,5π/3) and (5π/3,2π). All you need to do is to check that f(x) is positive in the first of those intervals.
I only considered Sin2x-Sinx=0 to find the roots.... um just because I recalled that as part of obtaining the answer to questions like this.
I wasn't really thinking f(x)=Sin2x-Sinx as a discrete function here, for me to obtain the answer through it or something.

I was just thinking about the 2 graphs Sin2x and Sinx and I was thinking that the regions where the Sin2x dipped below Sinx was the answer to this question.

I can't quite connect the consideration of f(x) here and the checking of whether f(x) is positive..............

Shall be grateful if you or someone else can shed some light on that

6. Hello Utterconfusion
Originally Posted by Utterconfusion
I only considered Sin2x-Sinx=0 to find the roots.... um just because I recalled that as part of obtaining the answer to questions like this.
I wasn't really thinking f(x)=Sin2x-Sinx as a discrete function here, for me to obtain the answer through it or something.

I was just thinking about the 2 graphs Sin2x and Sinx and I was thinking that the regions where the Sin2x dipped below Sinx was the answer to this question.

I can't quite connect the consideration of f(x) here and the checking of whether f(x) is positive..............

Shall be grateful if you or someone else can shed some light on that
You are quite right: the solution to the problem is equivalent to finding out where the graph of $\displaystyle y = \sin2x$ 'dips below' the graph of $\displaystyle y = \sin x$.

In the top diagram of the attached graphic, you'll see these two graphs on the same axes - and you can see where the pink (I hope it's pink on your monitor!) graph, $\displaystyle y = \sin2x$ dips below the blue graph, $\displaystyle y = \sin x$. (I've marked the angles in degrees, because it's easier to show their values that way on my spreadsheet graph.)

In the lower half of the graphic, I've plotted a single graph: $\displaystyle y = \sin2x - \sin x$. You'll see that this goes below the $\displaystyle x$-axis at exactly the same values as before, where the pink graph was below the blue graph.

It couldn't be simpler really: if $\displaystyle A < B$, then $\displaystyle A-B < 0$ simply because you're taking away the bigger number ($\displaystyle B$) from the smaller one ($\displaystyle A$), so you get a negative answer.

Don't make problems for yourself that aren't really there!

Hello UtterconfusionYou are quite right: the solution to the problem is equivalent to finding out where the graph of $\displaystyle y = \sin2x$ 'dips below' the graph of $\displaystyle y = \sin x$.

In the top diagram of the attached graphic, you'll see these two graphs on the same axes - and you can see where the pink (I hope it's pink on your monitor!) graph, $\displaystyle y = \sin2x$ dips below the blue graph, $\displaystyle y = \sin x$. (I've marked the angles in degrees, because it's easier to show their values that way on my spreadsheet graph.)

In the lower half of the graphic, I've plotted a single graph: $\displaystyle y = \sin2x - \sin x$. You'll see that this goes below the $\displaystyle x$-axis at exactly the same values as before, where the pink graph was below the blue graph.

It couldn't be simpler really: if $\displaystyle A < B$, then $\displaystyle A-B < 0$ simply because you're taking away the bigger number ($\displaystyle B$) from the smaller one ($\displaystyle A$), so you get a negative answer.

Don't make problems for yourself that aren't really there!

GRANDAD! I've finally got what the heck was goin on! THANKS A BUNCH! When I win the Nobel Price for Trigonometry, I'll mention you in my acceptance speech!@#$! =D Bless you & your grandchildren ! 8. Originally Posted by Utterconfusion GRANDAD! I've finally got what the heck was goin on! THANKS A BUNCH! When I win the Nobel Price for Trigonometry, I'll mention you in my acceptance speech!@#$! =D