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

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To solve the equation x*sin(x/2) - 18/pi = 0, users can utilize Newton's method of approximation, which requires an initial guess based on a table of x and f(x) values. By selecting x values like pi, 2pi, and 3pi, one can identify where f(x) changes sign, indicating a root exists between those points. The midpoint of these values serves as the starting point for Newton's method, which converges quickly if the initial guess is close to the actual solution. Alternatively, users can graph the function using a calculator like the TI-83 Plus to find the zero directly. Both Newton's method and the bisection method are viable options, with the latter being simpler but slower.
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 alot. 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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