Question

Find a real root of the equation 3x-cosx-1=0 using newton- raphson method

Answer 1

Answer:

Root : x= 0.607102

Step-by-step explanation:

In Newton- Raphson method, we use the formula to find the roots.

x_n = x_n - $\frac{f(x_n)}{f'(x_n)}$

f(x) = 3x-cosx-1

f'(x)= 3+sinx

Lets start with x_1 = 0.5

x_2 = 0.5 - $\frac{3(0.5)-Cos(0.5)-1}{3+sin(0.5)}$

      = 0.5 - (-0.108)

x_2     = 0.608

Next, x_3 =  0.608 - $\frac{3(0.608)-Cos(0.608)-1}{3+sin(0.608)}$

                = 0.608 - 0.000898

        x_3 = 0.607102

Hence, the real root is 0.607102. Continuing further. we get the values near to it..