# Trigonometric Inequality Solutions for x

• Physicsissuef
In summary, the student is trying to find the solution for x which is in the interval (-\pi, \pi) and they are not sure if it is x=\pi or if it is x=\frac{5\pi}{6}+\frac{-\pi}{2}. After graphing the function f(x) and y=\sqrt{3}/2 on the same grid, they see that x is in the interval (-\pi, \frac{5\pi}{6}+\frac{-\pi}{2}) and it is clear that the student's solution does not work.
Physicsissuef

## Homework Statement

Find the solutions for x:

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

## 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
$$x\in (\frac{5\pi}{6} , \frac{\pi}{3})$$

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.

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?

Physicsissuef said:
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
$$x\in (\frac{5\pi}{6} , \frac{-\pi}{2})$$

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.

## What is trigonometry sinus inequality?

Trigonometry sinus inequality is a mathematical concept that involves inequalities between trigonometric functions, specifically the sine function. The sine function is used to calculate the ratio of the opposite side to the hypotenuse in a right triangle.

## What is the importance of understanding trigonometry sinus inequality?

Understanding trigonometry sinus inequality is important for solving various mathematical problems involving triangles and angles. It is also used in many real-world applications, such as in navigation, engineering, and physics.

## How do you solve trigonometry sinus inequalities?

To solve a trigonometry sinus inequality, you first need to determine the range of values for the angle in question. Then, you can use the properties of the sine function and algebraic manipulation to solve the inequality and find the range of values for the angle.

## What are some common mistakes made when solving trigonometry sinus inequalities?

Some common mistakes made when solving trigonometry sinus inequalities include forgetting to consider the range of values for the angle, incorrectly applying the properties of the sine function, and making calculation errors.

## Can trigonometry sinus inequalities be applied to non-right triangles?

No, trigonometry sinus inequalities can only be applied to right triangles. The properties of the sine function and trigonometric ratios are specific to right triangles and cannot be applied to non-right triangles.

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