Calculus: Early Transcendentals, Chapter 4, 4.8, Section 4.8, Problem 17

3cos(x)=x+1
f(x)=x+1-3cos(x)=0
To solve using Newton's method apply the formula,
x_(n+1)=x_n-f(x_n)/(f'(x_n))
f'(x)=1+3sin(x)
Plug in f(x) and f'(x) in the formula,
x_(n+1)=x_n-(x_n+1-3cos(x_n))/(1+3sin(x_n))
See the attached graph to get the initial values of x.
When f(x)=0 , the values of x are near The curve has three x values x ~~ 0.8 , -1.8 , -3.6
Use these three values for finding the roots of the equation to six decimal places.
Let's solve for the first zero by initial value x_1=0.8
x_2=0.8-(0.8+1-3cos(0.8))/(1+3sin(0.8))
x2~~0.892041194
x_3~~0.889472276
x_4~~0.889470408
x_5~~0.889470408
Let's stop iteration as we have same decimal places
Now let's solve for the second zero by initial value x_1=-1.8,
x_2=(-1.8)-(-1.8+1-3cos(-1.8))/(1+3sin(-1.8))
x_2~~-1.861613881
x_3~~-1.8623648
x_4~~-1.862364929
x_5~~-1.862364929
Now let's solve for the third root by initial value x_1=-3.6
x_2=(-3.6)-(-3.6+1-3cos(-3.6))/(1+3sin(-3.6))
x_2~~-3.638785336
x_3~~-3.637958339
x_4~~-3.637957968
x_5~~-3.637957968
Roots of the equation to six decimal places are 0.889470 , -1.862365 , -3.637958