# Trigonometry sinus inequality

1. Apr 28, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

Find the solutions for x:

$$sinx<\frac{\sqrt{3}}{2}$$

2. Relevant equations

3. The attempt at a solution

Will the solution for x be
$$x\in (\frac{-\pi}{2} , \frac{\pi}{3})$$ ??
or
$$x\in (\frac{5\pi}{6} , \frac{-\pi}{2})$$ or
$$x\in (\frac{5\pi}{6} , \frac{\pi}{3})$$

2. Apr 28, 2008

### Tom Mattson

Staff Emeritus
1.) Take out a piece of graph paper.

2.) Graph the function $f(x)=\sin(x)$ over a few periods.

3.) On the same grid, graph the line $y=\sqrt{3}/2$.

The solution should be obvious after that.

3. Apr 28, 2008

### Physicsissuef

I see $$x\in (\frac{-\pi}{2} , \frac{\pi}{3})$$, but I am not sure, because all of this solutions are acceptable.

4. Apr 28, 2008

### Tom Mattson

Staff Emeritus
Yes, clearly there are infinitely many intervals on which the inequality is satisfied. The solution is the union of all of them.

5. Apr 28, 2008

### Physicsissuef

But if I write
$$x\in (\frac{-\pi}{2} +2k\pi , \frac{\pi}{3}+2k\pi)$$
It will be useless to make union of all of them, right?

6. Apr 28, 2008

### Tom Mattson

Staff Emeritus
I don't agree that those are the intervals that satisfy the inequality. Look at the graph again. You should see that one of the intervals for which $\sin(x)<\sqrt{3}/2$ is:

$$\left(\frac{2\pi}{3},\frac{7\pi}{3}\right)$$

That interval isn't captured by any value of $k$ in your solution.

What makes you say that? The problem statement was not restricted to some subset of the real line. If you want to correctly answer the question then you must express it as the union of all of the intervals on which the inequality is satisfied.

7. Apr 28, 2008

### Physicsissuef

They don't? I think they do. I draw trigonometric circular and they fit very well.

They are same with
$$x\in (\frac{5\pi}{6} , \frac{-\pi}{2})$$

8. Apr 28, 2008

### Tom Mattson

Staff Emeritus
If you draw the picture that I advised you to draw, then it is perfectly obvious that your solutions do not work.

Here is a simple counterexample. The inequality is satisfied for $x=\pi$. Now look closely at your solutions.

For $k=0$, the interval is:

$$\left(\frac{-\pi}{2},\frac{\pi}{3}\right )$$.

For $k=1$, the interval is:

$$\left(\frac{3\pi}{2},\frac{7\pi}{3}\right )$$.

Your solution misses $x=\pi$, and it is clearly wrong. Again: Draw the picture I advised you to draw, and it really ought to be clear.

9. Apr 29, 2008

### Physicsissuef

Yes, sorry. I realized that is $$\left(\frac{2\pi}{3},\frac{7\pi}{3}\right)$$

10. Apr 29, 2008

### BrendanH

As I am usually finicky, I just want to point at that, if you intend to include the constant 'k' in your solution, you should identify that it is an element of the natural numbers only.