



 
Hi Ming. I tried quickly sketching the two graphs y = sin(x) and y = x^{2}  x and I can immediately see two solutions exist: So I use guessandcheck: x = 0 works because sin(0) = 0^{2}  0 It looks like y = 1 is close. So for the second solution, I can only approximate. To do so, I can use "Newton's Method" to find successively more accurate values of x by starting with an x that is "close" to the solution. Newton's method says that if you have a function f(x) = 0, then you can start with an x that is close to a root and iterate x_{i} = x_{i1}  f(x)/f '(x) to get ever more precise. So f(x) in your case is x^{2}  x  sin(x), which makes f'(x) = 2x  1  cosx. If I start with x = 1.5, then it generates a better approximation for x: x = 1.5  [ 1.5^{2}  1.5  sin(1.5) ] / [ 2(1.5)  1  cos(1.5) ] Now iterate again: x = 1.628284745 [ 1.628284745^{2}  1.628284745 sin(1.628284745) ] / [ 2(1.628284745)  1  cos(1.628284745) ] And again: x = 1.617620013 [ 1.617620013^{2}  1.617620013 sin(1.617620013) ] / [ 2(1.617620013)  1  cos(1.617620013) ] And again: x = 1.617545290 [ 1.617545290^{2}  1.617545290 sin(1.617545290) ] / [ 2(1.617545290)  1  cos(1.617545290) ] which is such a tiny difference from the previous figure, I decide this is a good enough approximation. Cheers,  


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