Calculus: Early Transcendentals, Chapter 4, 4.8, Section 4.8, Problem 22

sin(x)=x^2-2
f(x)=x^2-2-sin(x)=0
f'(x)=2x-cos(x)
See the attached graph. From the graph, the curve of the function intersects the x-axis at ~~ -1.05 and 1.70. These can be used as initial approximations for further iteration.
x_(n+1)=x_n-((x_n)^2-2-sin(x_n))/(2x_n-cos(x_n))
For x_1=-1.05
x_2=-1.05-((-1.05)^2-2-sin(-1.05))/(2(-1.05)-cos(-1.05))
x_2=-1.05-(1.1025-2-(-0.867423225))/(-2.1-0.497571047)
x_2~~-1.061578807
carry out iteration until six digit decimal places are same
x_3~~-1.061549775
x_4~~-1.061549775
Now for x_1=1.70
x_2=1.7-(1.7^2-2-sin(1.7))/(2*1.7-cos(1.7))
x_2=1.7-(2.89-2-0.99166481)/(3.4-(-0.128844494))
x_2~~1.72880966
x_3~~1.728466368
x_4~~1.728466319
x_5~~1.728466319
Roots of the equation to six decimal places are -1.061550 , 1.728466