The Newton-Raphson's method uses the formula:

    xn+1=xn−f(xn)f'(xn)

So we need to change the equation into a function. This is done by moving all terms to one side:

    f(x)=sinx−x2

And we need the derivative:

    f'(x)=cosx−2x

The easiest way to iterate is to use a calculator. Enter f(x) into Y1 and f'(x) into Y2. Then enter a very short program that does this:

    A−Y1(A)Y2(A)→A

You need a starting value, x1. Since the question is asking for a positive root, we know that sin(x) has a maximum value of 1 at x=π2 and x2 is 1 at x=1. So it is safe to say that the root is less than 1 and therefore will be the starting value. We don't want a starting value close to 0 because 0 is also a root but not positive. So enter the following into your calculator,

    1→A

Then execute the program until you get the desired accuracy:

    0.8913959953
    0.8769848448
    0.8767262985
    0.8767262154
    0.8767262154

We get 3 digits of accuracy after 2 iterations, 7 after 3 iterations, and 10 after 4 iterations. So the answer converges very quickly for this root.
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