Use Newton's method to approximate the non-zero solution of the equation sin x-x^2 = 0.?

You have f(x) = sin x - x² and f'(x) = cos x - 2x is easy.

Newton's method says that if x_n is close enough to a root then:

x_(n+1) = x_n - f(x_n)/f'(x+n) = x_n - (sin x_n - x_n²)/(cos x_n - 2x_n)

...is closer. The only hitch here is finding a starting point that's "close enough". I'd start by approximating sin x by the first two terms of its power series:

sin x ~~ x - x²/2
sin x - x² ~~ x - (3/2)x² = x(1 - 3x/2)

That's 0 when x=0 or x=2/3. So, I'd start with x_0 = 2/3 and see if that converges on something nonzero.

x_1 = x_0 - (sin x_0 - x_0²)/(cos x_0 - 2x_0) ~~ 0.984
x_2 = x_1 - (sin x_1 - x_1²)/(cos x_0 - 2x_0) ~~ 0.888
x_3 = x_2 - (sin x_2 - x_2²)/(cos x_2 - 2x_2) ~~ 0.877

You don't need to keep a lot of digits at first. Once you have a few significant digits. though, the number of significant digits will usually double on each iteration.

x_4 ~~ 0.876726
x_5 ~~ 0.876726215395

The graph at the source link shows the approximate location of the zero to be x ≈ 0.8767. One iteration refines that result to
.. x ≈ 0.87672622

It took me years to get my head around Newton's method, and once I did I wondered how the helll it took me so long to learn! Perhaps a mental block. It really is quite easy, you just need to do the maths in the correct sequence!