Solve sin(y/2)=y/4: Step-by-Step Guide

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The equation sin(y/2) = y/4 has no algebraic solution, but three distinct solutions exist, including y = 0. The simplest method to find these solutions is by graphing both functions and identifying their intersection points. An alternative approach is the bisection method, which involves narrowing down the interval where the two functions intersect through iterative calculations. This method requires checking values between known bounds until the desired accuracy is achieved. The discussion suggests that while algebraic methods may be attempted, they are likely not effective for this equation.
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Please show the steps
 
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There is no "algebraic" way to solve that equation. (There are, in fact, three distinct solutions. Here's one for free: x= 0!)

Simplest way: graph y= sin(x/2) and graph y= x/4 on the same coordinates. The solution is the x coordinate of the points where the two graphs cross. On my TI85 calculator, that's quick and simple. It even let's me "zoom" in on a point of intersection.

Almost as simple but tedious is 'bisection": going by steps of 1, I see that sin(3/2)= .997 and 3/4= .75 so sin(x/2) is larger. But sin(4/2)= .909 while 4/4= 1 so now x/4 is larger. Since sin(x/2) and x/4 are continuous functions, that means that they must be equal somewhere between 3 and 4. We don't know where so 1/2 way between is as good as any: sin(3.5/2)= .984 and 3.5/4= .875. sin(x/2) is larger there so they must be equal somewhere between 3.5 and 4. Half way between again is 3.75. sin(3.75/2)= .984 and 3.75/4= .934. sin(x/2) is still larger so we try half way between 3.75 and 4, 3.875. sin(3.875/2)= .9335 (I added an addtional decimal place because these are all the same to 3 places) and 3.875/4= .9687. Now x/4 is larger so there must be a solution between 3.75 (the last value at which sin(x/2) was larger) and 3.875. Half way between them is 3.8125. Continue this until you have the desired accuracy. Once you get that root, it should be obvious what the third solution is.
 
I'm sure you could try the half-angle formula and then solve algebraically from there. I'm not sure if it would work out though.
 
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