# How do you solve sin x - 2x = 0 ?

1. Jul 22, 2014

### johann1301

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

solve sin x - 2x = 0

3. The attempt at a solution

I can see that zero is a solution, but i dont know how to actually solve it in a general way. Can somebody give me a hint?

2. Jul 22, 2014

### Char. Limit

Well, sin(x) - 2x = 0 implies that sin(x) = 2x. From there, you have two options:

1. Graph it, and you'll be able to see just how many solutions this equation has.

2. If you're looking for an analytical proof, then considering that sin(x) is limited in range to [-1,1], you know you only have to consider solutions within x = [-1/2, 1/2] due to the range of 2x, and within that range, you can take a third-order Taylor approximation of sin(x) to create a cubic that you can look at.

3. Jul 22, 2014

### johann1301

Thank you!

4. Jul 22, 2014

### HakimPhilo

There's the obvious solution $x=0$, do you know how to prove that it is the only solution?

5. Jul 22, 2014

### johann1301

Use the second method that "Char. Limit" proposed. Or perhaps you know another way? I sure dont...

6. Jul 22, 2014

### Char. Limit

Obviously, if you use the method that I proposed, you'll also want to calculate the /error/ in that approximation, but it's not too difficult to see the remaining steps.

7. Jul 22, 2014

### HakimPhilo

Consider $f(x)=\sin x-2x$.
1. Prove that $f$ is strictly decreasing.
2. Find two numbers $\alpha$ and $\beta$ in $\Bbb R$ such that $f(\alpha)\gt0$ and $f(\beta)\lt0$.
Now use the intermediate value theorem and you're done.

8. Jul 22, 2014

### johann1301

Thanks, but its a bit to complex for me.