Question

find the real root of the equation x^2+4sinx=0 by using newton raphson's method​

Answer 1

Answer:

By seeing the equation itself we can say two things

1) It has infinite roots

2) One among the root is x = 0

because at x=0 f(x)=0

But if you want to find one more root by newton raphson,

Choose the value x0

Put x=pi(3.14) i.e., 180 degree

f(x)=3.14^2 + 4sin180 (Set your calculator in degree mode)

f(x)=9.8596+0

This is not close to 0 So take some other value say x= - 11pi / 18 i.e., -110 degree

(Conversion of pi to degree (value*180/3.14) )

(Conversion of degree to pi (value * 3.14/180) )

f(x) = (-11 *3.14 / 18)^2 + 4sin(-110) (Set your calculator in degree mode)

f(x) = -.08 (Here I am just taking two digit accuracy.If you want you may take more number of digits also)

It is close to 0 So take xo=-11pi / 8

Now Use the formula x1= x0- f(x0) / f’(x0)

here f’(x) = 2x +4cosx

So f’(-11pi/18) = —5.21

So x1 = -11pi/18 - (-.0

Answer 2

Answer:

Step-by-step explanation: