Use Newton’s method to find all roots of the equation correct to six decimal places. 3 cos(x) = x+1?

We have:

# 3cosx=x+1 #

Let:

# f(x) = 3cosx-x-1 #

Our aim is to solve #f(x)=0#. First let us look at the graph:
graph{3cosx-x-1 [-10, 10, -5, 10]}

To find a solution numerically, using Newton-Rhapson method we use the following iterative sequence

# { (x_1,=x_0), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :} #

Therefore we need the derivative:

# \ \ \ \ \ \ \f'(x) = -3sinx-1 #

We can see from the graph there are three solutions. The root we find will depend upon our initial approximation #x_0#, and this value will require a little trial and error.

Then using excel working to 8dp we can tabulate the iterations as follows:

Initial Value: #x_0 = 1#

Initial Value: #x_0 = -1#

Initial Value: #x_0 = -3#

We could equally use a modern scientific graphing calculator as most new calculators have an " Ans " button that allows the last calculated result to be used as the input of an iterated expression.

And iterating until we get convergence, we conclude that the solution (to 8dp) are:

# x=0.88947041,-1.86236493,-3.63795797 #

So Rounding to #6dp# we have

# x=0.889470,-1.862365,-3.637958 #