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

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In summary, the conversation discusses solving the equation sin(x) - 2x = 0 and provides two methods for finding the solutions. The first method involves graphing the equation and the second method involves using a third-order Taylor approximation. It is mentioned that the solution x=0 can be easily proven using the second method. The conversation also mentions using the intermediate value theorem to prove the existence of solutions.
  • #1
johann1301
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Homework Statement



solve sin x - 2x = 0

The Attempt at a Solution



I can see that zero is a solution, but i don't know how to actually solve it in a general way. Can somebody give me a hint?
 
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  • #2
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.
 
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  • #3
Thank you!
 
  • #4
There's the obvious solution ##x=0##, do you know how to prove that it is the only solution?
 
  • #5
Use the second method that "Char. Limit" proposed. Or perhaps you know another way? I sure dont...
 
  • #6
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
johann1301 said:
Use the second method that "Char. Limit" proposed. Or perhaps you know another way? I sure dont...

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
Thanks, but its a bit to complex for me.
 

1. What is the first step in solving sin x - 2x = 0?

The first step in solving this equation is to isolate the sin x term by adding 2x to both sides of the equation, resulting in sin x = 2x.

2. How do you solve for x when both sides of the equation have trigonometric functions?

To solve for x in this type of equation, you can use a graphing calculator or a table of values to find the intersections between sin x and 2x. This will give you approximate values for x, which can then be refined using algebraic techniques.

3. Can this equation have multiple solutions?

Yes, this equation can have multiple solutions. Since the sine function is periodic, there are infinitely many values of x that satisfy the equation sin x = 2x. However, when solving for x, we typically only consider solutions within a specific interval, such as 0 to 2π.

4. Are there any special techniques for solving equations involving trigonometric functions?

Yes, there are several techniques for solving these types of equations, such as using the unit circle, trigonometric identities, and inverse trigonometric functions. It is important to have a good understanding of these concepts when working with trigonometric equations.

5. Can this equation be solved algebraically?

No, this equation cannot be solved algebraically. Because of the periodic nature of the sine function, it is not possible to isolate x on one side of the equation using algebraic techniques. Instead, we must use a combination of algebra and trigonometry to find approximate solutions.

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