Solving this equation with both x, sinx and cosx

• lo2
In summary: So g(x) is positive on that interval. On the interval, (...,pi/2), g(x) is increasing and on (pi/2,...), g(x) is decreasing. g(pi/2)=0, so g(x) is never negative. Therefore, g(x) has no zeros on the interval, (0,pi/2)\ . On the interval, (pi/2,pi), g(x) is negative and decreasing. It's not too difficult to show that g(x) has exactly one zero on the interval, (pi/2,pi
lo2

Homework Statement

I have this function:

$f(x) = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}}$

For all $x \in R$ where $x \neq n \pi, n \in Z$

And then I have to solve this equation f(x)=0:

$f(x) = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}}$

Where I have to show that it has no solutions in the interval 0 < x < Pi and that it has got one solution in the interval Pi < x < 2Pi, and then I have to approximate that solution using Maple.

The Attempt at a Solution

You get this equation:

$0 = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}} ⇔ \frac{1}{x}=\frac{\cos{(x)}}{\sin{(x)}}$

Or you can rewrite it to:

$0 = \frac{\sin{(x)}-x \cos{(x)}}{x \sin{(x)}}$

And then just show that the nominator will never be 0.

But anyhow I cannot really see how to make proper progess, so how can I solve this? And is it true that an equation with both x and sin(x) (or cos(x)) will not have an algebraic solution?

lo2 said:

Homework Statement

I have this function:

$f(x) = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}}$

For all $x \in R$ where $x \neq n \pi, n \in Z$

And then I have to solve this equation f(x)=0:

$f(x) = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}}$

Where I have to show that it has no solutions in the interval 0 < x < Pi and that it has got one solution in the interval Pi < x < 2Pi, and then I have to approximate that solution using Maple.

The Attempt at a Solution

You get this equation:

$0 = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}} ⇔ \frac{1}{x}=\frac{\cos{(x)}}{\sin{(x)}}$

Or you can rewrite it to:

$0 = \frac{\sin{(x)}-x \cos{(x)}}{x \sin{(x)}}$

And then just show that the nominator will never be 0.

But anyhow I cannot really see how to make proper progess, so how can I solve this? And is it true that an equation with both x and sin(x) (or cos(x)) will not have an algebraic solution?

When confronted with a big problem, try and learn to scale it down and just focus on a small part or a simplified version of it. So, just for now, assume you only had one problem to solve:

For $f(x)=1/x-\frac{\cos(x)}{\sin(x)}$, show that it doesn't have a zero in the interval $(0,\pi)$

Now let's cheat a little bit. Huh? Nothing's wrong with that. I mean just plot them for heaven's sake. Yeah, I did and I can see 1/x and cot(x) don't. Now when you combined them you determined that in order for it to have a solution there, the expression:

$$\sin(x)-x\cos(x)$$

would have to have a zero there. Alright, I'm not proud. Plot those two also. Yeah, they both start at the origin and one is . . . ALWAYS . . . below the other one. Now, for f(x)=sin(x) and g(x)=x cos(x) with g(x) always being lower than the other, what analytic property of g(x) would ensure it to always be lower if they start off at the same point?

There is, in general, no "algebraic" method to get an exact value of an equation in which the uknown, x, appears both inside transcendental functions (such as sin(x) and cos(x)) and outside. I suppose you might be able to write sine and cosine as complex exponentials ($sin(x)= (e^{ix}- e^{-ix})/2i$ and $cos(x)= (e^{ix}+ e^{-ix})/2$) and use a complex version of the "Lambert W function" (defined as the inverse function to $f(x)= xe^x$) but that would be extremely complex. Best to use a numerical method like "Newton's method".

But is there an analytical way of showing that the equation has no solution in the interval ]0;Pi[ and has got one solution ]Pi;2Pi[?

What's the first tool that comes to mind when you are asked to show something analytically? Have you tried using it?

D H said:
What's the first tool that comes to mind when you are asked to show something analytically? Have you tried using it?

Ok well I do not know if what I mean can be called analytical.

I just wanted to be able to reach an expression from which I can show that there is no solution and one solution respectively.

You posted this in the "Calculus & Beyond" section. What is one of the first things you learn when you are taking calculus, just after learning about limits?

lo2 said:
Ok well I do not know if what I mean can be called analytical.

I just wanted to be able to reach an expression from which I can show that there is no solution and one solution respectively.

Ok, let's do an easier one. Consider:

$$f(x)=x^2$$

$$g(x)=1/2 x^2$$

They too start at zero and never cross so the expression $x^2-1/2 x^2$ is never zero for x ne 0. Note also that the plot of $1/2 x^2$ is always below the plot for $x^2$. Now, they both start at zero and one way (the simplest) for g(x) to always be lower than the other is for the rate at which g(x) is changing to be always less than the rate at which f(x) is changing. Now, how can you analytically represent that rate phrase in terms of Calculus?

I don't think I see the following approach used in this thread so far, but I have a history of missing stuff that's been previously posted in a thread.

Finding the zeros of $\displaystyle f(x)=\frac{1}{x}-\frac{\cos(x)}{\sin(x)}$ is equivalent to solving $\displaystyle \frac{1}{x}=\cot(x)\ .$

But that's equivalent to solving $\displaystyle x=\tan(x)\,,$ for all x in the domain of $f(x)\ .$

The solutions to the equation $\displaystyle x=\tan(x)$ are the same as the zeros of the function $\displaystyle g(x)=\tan(x)-x\ .$

It's not too difficult to show that $g(x)$ has no zeros on the interval, (0, π).

lo2 said:

But is there an analytical way of showing that the equation has no solution in the interval ]0;Pi[ and has got one solution ]Pi;2Pi[?

To show that g(x) = sin(x) - x*cos(x) has no root in the first interval, look at its derivative g'(x) = cos(x) - cos(x) + x*sin(x) = x*sin(x). This is > 0 in (0,π), which means that g is strictly increasing throughout the whole interval. Since g(x) --> 0 as x --> 0+, that means that the graph of y = g(x) cannot cross zero within the interval. As for the interval (π,2π): g starts off at g(π) = +π, and ends at g(2π) = -2π, so switches from + to - in the interval. Therefore, it has a root. Within the interval (π,2π) its derivative is < 0, so it is strictly decreasing throughout that interval, and that makes the root unique.

RGV

Last edited:
Are you allowed to plot a graph of y = x and y = tan (x) on the same plot, or is that against the rule? On such a graph, you will see right away that the two plots don't cross over each other until the third quadrant.

Ah ok I think I got it now!

I just showed that in the first interval ]0;Pi[ it starts out at f(x) > 0 and that it then heads toward infinity and since it is both only growing and continuous it will not cross the x axis.

And in the other interval ]Pi;2Pi[ it starts out coming from -infinity and goes toward infinity and since it also only growing and continuous in this interval, it has to cross the x-axis just once.

Btw does anyone how to approximate this solution in Maple? Where you only get this one solution.

lo2 said:
Ah ok I think I got it now!

I just showed that in the first interval ]0;Pi[ it starts out at f(x) > 0 and that it then heads toward infinity and since it is both only growing and continuous it will not cross the x axis.

And in the other interval ]Pi;2Pi[ it starts out coming from -infinity and goes toward infinity and since it also only growing and continuous in this interval, it has to cross the x-axis just once.

Btw does anyone how to approximate this solution in Maple? Where you only get this one solution.

In good English, are you asking how to get Maple to solve the equation in the interval (π, 2π)?
It's easy:

> eq:=1/x - cot(x)=0;
eq := 1/x - cot(x) = 0

> fsolve(eq,x=Pi..2*Pi);
4.493409458
If you want more accuracy, just increase the Digits count:
> Digits:=20;
Digits := 20

> fsolve(eq,x=Pi..2*Pi);
4.4934094579090641753

RGV

1. How do I solve an equation with both x, sinx and cosx?

To solve an equation with both x, sinx and cosx, you need to use trigonometric identities and algebraic techniques to manipulate the equation into a form where you can isolate the variable x.

2. Can I use a calculator to solve this type of equation?

Yes, you can use a calculator to help you solve the equation, but it is important to understand the steps involved in solving it by hand first.

3. What are some common trigonometric identities that I can use?

Some common trigonometric identities that can be used to solve equations with x, sinx, and cosx include the Pythagorean identities, double angle identities, and sum-to-product identities.

4. Are there any specific strategies for solving these types of equations?

Yes, there are a few strategies that can be helpful when solving equations with x, sinx, and cosx. These include factoring, substitution, and using the unit circle to simplify trigonometric expressions.

5. Is it possible to have multiple solutions for an equation with x, sinx, and cosx?

Yes, it is possible to have multiple solutions for an equation with x, sinx, and cosx. This is because trigonometric functions are periodic and have repeating solutions. It is important to check for extraneous solutions when solving these types of equations.

• Calculus and Beyond Homework Help
Replies
3
Views
343
• Calculus and Beyond Homework Help
Replies
2
Views
914
• Calculus and Beyond Homework Help
Replies
11
Views
359
• Calculus and Beyond Homework Help
Replies
3
Views
558
• Calculus and Beyond Homework Help
Replies
7
Views
706
• Calculus and Beyond Homework Help
Replies
18
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
176
• Calculus and Beyond Homework Help
Replies
6
Views
548
• Calculus and Beyond Homework Help
Replies
21
Views
839
• Calculus and Beyond Homework Help
Replies
5
Views
1K