Help with trigonometry inequation

[Johulus]

I don't know how should I solve this:

\(\displaystyle x\, \cdot\, \sin\left(x\right)\, +\, 1\, >\, x\, +\, \sin\left(x\right) \)

I tried something like:

\(\displaystyle x\, \cdot\, \sin\left(x\right)\, +\, 1\, >\, x\, +\, \sin\left(x\right) \\ x\, \cdot\, \sin\left(x\right)\, -\, \sin\left(x\right)\, >\, x\, -\, 1 \\ \sin\left(x\right)(x\, -\, 1)\, -\, (x\, -\, 1)\, >\, 0 \\ (x\, -\, 1)(\sin\left(x\right)\, -\, 1)\, >\, 0\)

But, I don't know what to do then.... I've tried something but then it came out completely different than the result for this task that I have in the book. So, I am not really sure what to do with this. Also, \(\displaystyle (\sin\left(x\right)\, -\, 1)\) is always smaller than 0, except if being 0 but that is out of the question here, so for the whole expression to be larger than 0 I guess \(\displaystyle (x\, -\, 1)\) should be also smaller than 0. But I'm confused and I can't get to the solution that is provided in my book for that task.

This is what is provided as the solution for this task:
\(\displaystyle -\dfrac{3\pi}{2}\, -\, k\, \cdot\, 2\pi\, <\, x\, <\, \dfrac{\pi}{2}\, +\, k\, \cdot\, 2\pi\, , k\, \in\, N_{0} \)

 

[stapel]

I don't know how should I solve this:

\(\displaystyle x\, \cdot\, \sin\left(x\right)\, +\, 1\, >\, x\, +\, \sin\left(x\right) \)

I tried something like:

\(\displaystyle x\, \cdot\, \sin\left(x\right)\, +\, 1\, >\, x\, +\, \sin\left(x\right) \\ x\, \cdot\, \sin\left(x\right)\, -\, \sin\left(x\right)\, >\, x\, -\, 1 \\ \sin\left(x\right)(x\, -\, 1)\, -\, (x\, -\, 1)\, >\, 0 \\ (x\, -\, 1)(\sin\left(x\right)\, -\, 1)\, >\, 0\)

But, I don't know what to do then.... I've tried something but then it came out completely different than the result for this task that I have in the book. So, I am not really sure what to do with this. Also, \(\displaystyle (\sin\left(x\right)\, -\, 1)\) is always smaller than 0, except if being 0 but that is out of the question here, so for the whole expression to be larger than 0 I guess \(\displaystyle (x\, -\, 1)\) should be also smaller than 0. But I'm confused and I can't get to the solution that is provided in my book for that task.

This is what is provided as the solution for this task:
\(\displaystyle -\dfrac{3\pi}{2}\, -\, k\, \cdot\, 2\pi\, <\, x\, <\, \dfrac{\pi}{2}\, +\, k\, \cdot\, 2\pi\, , k\, \in\, N_{0} \)

Thank you for showing your efforts, and for providing the expected solution. I think you've made a good start. Now you need to take it a little further. You have:

. . . . .\(\displaystyle x\, \sin\left(x\right)\, +\, 1\, >\, x\, +\, \sin\left(x\right) \)

You've rearranged as:

. . . . .\(\displaystyle (x\, -\, 1)(\sin\left(x\right)\, -\, 1)\, >\, 0\)

This product is positive when its two factors have the same sign. So x - 1 > 0 and also sin(x) - 1 > 0, or else x - 1 < 0 and also sin(x) - 1 < 0. Consider the two cases separately:

x - 1 > 0 and sin(x) - 1 > 0:

In this case, the second inequality says that sin(x) > 1. Is this ever true? So is there any solution in this case?

x - 1 < 0 and sin(x) - 1 < 0:

In this case, the second inequality says that sin(x) < 1, which is always true, except for where sin(x) = 1. So find the x-values for which sin(x) = 1, and note that these will be excluded from the solution.

The first inequality says that x < 1. Note that this appears to conflict with the expected solution, since, for instance, k = 1 gives an interval for x of -(7/2)pi < x < (5/2)pi. This would allow for a value of x of, say, x = 2pi. But this evaluates as:

. . . . .\(\displaystyle (2\pi)\sin(2\pi)\, +\, 1\, =\, 0\, +\, 1\, =\, 1\,<\, 6\, <\, 2\pi\, =\, 2\pi \, +\, 0\, =\, 2\pi\, +\, \sin(2\pi)\)

Unless I'm missing something (which has been known to happen), the expected solution must be incorrect (or else it's the solution to some other exercise, and the question is what's wrong). Try continuing what you'd started, and remember to check with a graph.

 

[Ishuda]

Starting at
[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]0[/FONT]
we have either both factors must be positive or both factors must be negative.

Let's look at the positive one first. If both factors are positive, then
[FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]0[/FONT]
or
[FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]1[/FONT]
which is clearly impossible for real x. Thus, if there are to be solutions, both factors must be negative.

If both factors are negative, then
x < 1
and
[FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]<[/FONT][FONT=MathJax_Main]1[/FONT]
So the answer is, the inequality is true when both x and sin(x) are less than one. Well sin(x) is always less than one except when it is equal to one. Where is that? Just take that region (those regions) out of x<1.

Note that is not quite what the book provides if you have copied it correctly.

 

[ClaireWu]

Starting at
[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]0[/FONT]
we have either both factors must be positive or both factors must be negative.

Let's look at the positive one first. If both factors are positive, then
[FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]0[/FONT]
or
[FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]1[/FONT]
which is clearly impossible for real x. Thus, if there are to be solutions, both factors must be negative.

If both factors are negative, then
x < 1
and
[FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]<[/FONT][FONT=MathJax_Main]1[/FONT]
So the answer is, the inequality is true when both x and sin(x) are less than one. Well sin(x) is always less than one except when it is equal to one. Where is that? Just take that region (those regions) out of x<1.

Note that is not quite what the book provides if you have copied it correctly.

So x<1 and sin(x)<1 where x is in radians

Looking at sine curve y=sin(x)...for sin(x)<1, x can take all values except pi/2 and 5pi/2 and so on

so for x<1 and sin(x)<1...the answer should be x<1

(I hope I am right)

 

[Ishuda]

So x<1 and sin(x)<1 where x is in radians

Looking at sine curve y=sin(x)...for sin(x)<1, x can take all values except pi/2 and 5pi/2 and so on

so for x<1 and sin(x)<1...the answer should be x<1

(I hope I am right)

Following what I said,
sin(x) = 1 at x = \(\displaystyle \frac{\pi}{2}\) + 2n\(\displaystyle \pi\)
where n belongs to the (negative and non-negative) integers.
Thus sin(x) is less than one in any open \(\displaystyle 2\pi\) interval:
\(\displaystyle -\frac{3\pi}{2} + 2n\pi\) < x < \(\displaystyle \frac{-3\pi}{2}\) + 2(n+1)\(\displaystyle \pi\) = \(\displaystyle \frac{\pi}{2}\) + 2n\(\displaystyle \pi\)
We could have chose any value for that \(\displaystyle -\frac{3\pi}{2}\) as long as it was \(\displaystyle \frac{\pi}{2} + 2m\pi\) for some m. The reason for the choice made is because I already know the answer and that choice works out nicely. I would get the same answer but would just have to manipulate formulas more to get the desired answer.

Now x must be less than 1 also so
\(\displaystyle \frac{pi}{2}\) + 2n\(\displaystyle \pi \le 1\)
so
\(\displaystyle n \le \frac{1-\frac{\pi}{2}}{2\pi}\)
or
n must be negative and we have
\(\displaystyle -\frac{3\pi}{2} - 2n\pi < x < \frac{\pi}{2} - 2n\pi\); n = 1, 2, 3, ...

But what about that n=0 region. Well, if cut the upper limit off at x=1 we can also use that, so we need to add the interval
\(\displaystyle -\frac{3\pi}{2} < x < 1\)

Now you see what I mean by the answer is not quite what was given.

 

[ClaireWu]

Following what I said,
sin(x) = 1 at x = \(\displaystyle \frac{\pi}{2}\) + 2n\(\displaystyle \pi\)
where n belongs to the (negative and non-negative) integers.
Thus sin(x) is less than one in any open \(\displaystyle 2\pi\) interval:
\(\displaystyle -\frac{3\pi}{2} + 2n\pi\) < x < \(\displaystyle \frac{-3\pi}{2}\) + 2(n+1)\(\displaystyle \pi\) = \(\displaystyle \frac{\pi}{2}\) + 2n\(\displaystyle \pi\)
We could have chose any value for that \(\displaystyle -\frac{3\pi}{2}\) as long as it was \(\displaystyle \frac{\pi}{2} + 2m\pi\) for some m. The reason for the choice made is because I already know the answer and that choice works out nicely. I would get the same answer but would just have to manipulate formulas more to get the desired answer.

Now x must be less than 1 also so
\(\displaystyle \frac{pi}{2}\) + 2n\(\displaystyle \pi \le 1\)
so
\(\displaystyle n \le \frac{1-\frac{\pi}{2}}{2\pi}\)
or
n must be negative and we have
\(\displaystyle -\frac{3\pi}{2} - 2n\pi < x < \frac{\pi}{2} - 2n\pi\); n = 1, 2, 3, ...

But what about that n=0 region. Well, if cut the upper limit off at x=1 we can also use that, so we need to add the interval
\(\displaystyle -\frac{3\pi}{2} < x < 1\)

Now you see what I mean by the answer is not quite what was given.

Yes i see what you mean...thank you