Question

use newton-raphson method to find a root of the equation xsinx+cosx=0​

Answer 1

Answer:

x= 75.3849

Step-by-step explanation:

The Newton-Raphson method is an iterative method to find roots. It requires a function set to zero and its first derivative.

xn+1=xn−f(x)/f′(x)

f(x)=xsin(x)+cos(x) function set to zero

f′(x)=xcos(x)+cos(x)−sin(x) first derivative

xn+1=xn−xnsin(xn)+cos(xn)/xncos(xn)+cos(xn)−sin(xn)

Compute the Newton-Raphson method steps starting with x0=5.

x1=5−5sin(5)+cos(5)/5cos(5)+cos(5)−sin(5)

x1≈4.75688108

x2=x1−x1sin(x1)+cos(x1)/x1cos(x1)+cos(x1)−sin(x1)

…

The reader can continue to find the real-valued root to 4 decimal places.

Newtons method xsin(x) + cos(x) = 0

x6≈75.3849