How do you get rid of the SIN() function to solve for x?

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SUMMARY

The discussion focuses on solving the equation x*sin(x/2) - 18/pi = 0, specifically on eliminating the SIN() function. Participants recommend using Newton's method of approximation, which requires finding an initial starting point, x0, by evaluating f(x) = x*sin(x/2) - 18/pi at various x values. A change in sign indicates the presence of a root between those x values. Additionally, users suggest using a TI-83 Plus calculator to graph the function and utilize its 'zero' function for finding solutions.

PREREQUISITES
  • Understanding of trigonometric functions, specifically sine.
  • Familiarity with Newton's method of approximation.
  • Basic knowledge of graphing techniques using a TI-83 Plus calculator.
  • Concept of function evaluation and sign changes in root-finding methods.
NEXT STEPS
  • Learn the detailed steps of Newton's method for root finding.
  • Explore the bi-section method as an alternative to Newton's method.
  • Investigate how to use the 'zero' function on a TI-83 Plus calculator.
  • Study periodic functions and their implications on root-finding techniques.
USEFUL FOR

Mathematicians, engineering students, and anyone involved in numerical analysis or solving trigonometric equations will benefit from this discussion.

r3dxP
i have x*sin(x/2) - 18/pi =0
and i need to solve for x..
how in the heck can i get rid of that sin() function? very confused.. any help will be appreicated..!
btw.. i need the answer ASAP! thanks a lot guys
 
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Can you use Newton's method of approximation ?
 
i guess you can, could u explain how i would do that? and give me the answer for x=? thanks
 
r3dxP said:
i guess you can, could u explain how i would do that? and give me the answer for x=? thanks
"[URL a good description[/URL] and an example of Newton's method of approximation.
You will have to find the initial starting point, x0 -- the firstapproximation -- yourself.

To do so, make up a table of x and f(x) values, where f(x) = x*sin(x/2) - 18/pi.
Use values of x like: pi, 2pi, 3pi, ... until you get a change in the sign of f(x). That means that the curve, f(x), has crossed the x-axis, and therefore has been, at some point, equal to zero between the two x-values that gave the change of sign.

Take half-way between the two x-values as your first approximation, x0. You can get a closer approximation to x0 if you are familiar with interpolation. But that might be a bit of overkill at this point.
 
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thanks a lot. i just figured out that i can just graph it with ti83plus, then go2 calc and use the 'zero' function :) thanks for the help
 
Even though you have the solution from the TI, here are additional comments that might be useful. Since this function is periodic, it will cross y=0 many times (see thumbnail). Newton's method works extremely well and converges very quickly on the solution, provided the starting guess of the solution is sufficiently close. Another option is bi-section, which is computationally much simpler, but takes longer to converge to the solution. Either method can easily be done w/ Excel.
 

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