# Trigonometry sinus inequality

1. Homework Statement

Find the solutions for x:

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

2. Homework 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
$$x\in (\frac{5\pi}{6} , \frac{\pi}{3})$$

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Tom Mattson
Staff Emeritus
Gold Member
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.

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

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

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?

Tom Mattson
Staff Emeritus
Gold Member
But if I write
$$x\in (\frac{-\pi}{2} +2k\pi , \frac{\pi}{3}+2k\pi)$$
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.

It will be useless to make union of all of them, right?
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.

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

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

Tom Mattson
Staff Emeritus
Gold Member
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.

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

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.