Using newton's method find the positive root of 6 sin x = x^2

  Newtons method x(i+1) = x(i) - f(x(i)/f '(x(i)), where f(x) is ideally 0. We do this until out f(x(i))=0. i is just the iteration that you perform. 

First get one side to equal zero. 
sin(x) = x2. 
sin(x) - x2 = 0 Set this to f(x) 
f(x) = sin(x) - x2 = 0. Find f ' (x) 
f ' (x) = cos(x) - 2x 

We pick a starting guess. We have to be careful about where we start. We generally pick a starting guess near a known solution. Graph the two values and observe that the only intersections are zero and approximately 1. 

Make a table with i, y(i), f(y(i)), f ' (y(i)) to keep you organized. When you get a number close to zero for f(y(i)) you have found a solution. For this particular problem, you can terminate your iterations when the sixth decimal place stops changing. 

Here are some of the iterations when we use 1 as a starting guess: 
1 
0.89139600 
0.876984845 
0.876726298 
0.876726215 
0.876726215 
0.876726215 
0.876726215 

Note that the sixth decimal point stops changing. 
Also note if we use -2 as a starting point, we get: 
-2 
-0.63016223 
-0.153248303 
-0.017213911 
-0.000284854 
-8.11E-08 
-6.58E-15 
-4.26E-29 
0 
0 
0 
0 

So always determine visually, if possible, the correct starting point. 

BTW, MATLAB code is as follows: 
y(1)=1 
for i =1:30 
x(i+1)=x(i) - (sin(x(i))-x(i)^2)/(cos(x(i))-2*x(i)); 
end 
x; 
zero=sin(x)-x(i).^2;

The answer is lol i dont know

Agc and bgd are vertical angles