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Solving $2x - \sin 2x = \pi/2$ for $0 < x < \pi/2$

For fun, I was trying to solve this problem without doing calculus. After dinking around with it for a while, I came across the following term and don't quite know how to solve it without guessing:

$a - \sin{(a)} = \frac{\pi}{2}$

Where $a$ will be the angle of the chord that will allow me to solve for the height - in theory :)

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    By the way, for future reference (or for immediate reference if you'd like), a better way to post "re-opening requests" is to do so over at http://meta.math.stackexchange.com/ When you do, please ask a new question requesting re-opening, with a link to the closed question, as well as why you think the question should be re-opened (for example, in this case you will need to explain why this particular question is not, in fact, a duplicate of the linked one). You will get many more views that way.2011-07-01

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You can rewrite the expression as $\sin(a) =a - \frac{\pi}{2}$ so that each side of the equality is a function of $a$, each of which can be graphed (on the same graph!).

The LHS can be graphed as $f(a) = \sin(a)$ the graph of which you should know well,

and the RHS can be graphed as $g(a) = a - \frac{\pi}{2}$ Clearly, the graph of $g(a)$ is simply a line, with y-intercept $\left(-\frac {\pi}{2}, 0\right)$ and slope $= 1$.

You can use your graphs, then, to approximate any potential solutions; that is, find any point(s) of intersection. From that, you can probably "trouble shoot" with your calculator to make this approximation more precise.

I'll include graphs below:

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enter image description here

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    @Michael Dorgan: It will definitely not turn into $4R/3\pi$. There is a classical way of solving the center of mass problem without calculus. If you are curious about it, you can post a question (without the $\sin\theta$ stuff) or send me a message. Posting is better, since you may get a number of different suggestions.2011-07-01