Newton's method

Let f(x) = (cos(x))^2 - 3*x

Show that abs(f''(x)/f'(y))<=B, for x,y exist in [0,1], B is constant>0
Hence or otherwise deduce that Newton's method will converge.

I found that
abs(f''(x)/f'(y)) = abs( (2(sin(x)^2) - 2*(cos(x)^2)) / (-2cosy*siny - 3))

I can deduce that abs(2(sin(x)^2) - 2*(cos(x)^2)) is <=2

and that abs(-2cosy*siny - 3)>=3

so abs( (2(sin(x)^2) - 2*(cos(x)^2)) / (-2cosy*siny - 3))<=2/3

and so B=1? Am I correct??

Thank you very much!!!

Couldn't edit my post, but I can now see that B>0 as a positive constant can be a fraction so is B=2/3?

And after that how can I show that Newton's method converges? If the condition abs(f''(x)/f'(y))<=B, for x,y exist in [0,1] is satisfied then don't we automatically know that the method converges?

Thank you very much!!!

Can someone help me please? :) Thanks again!