Question

find the real root of the the following equation by using bisection cosx-3x+1=0​

Answer 1

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

X_n = X_n f'(an)

f(x) = 3x-cosx-1

f'(x)= 3+sinx

Lets start with x_1 0.5

X_2 = 0.5 - 3(0.5)-Cos(0.5)-1

3+sin(0.5)

= 0.5-(-0.108)

= 0.608

X_2

Next, x_3 = 060830608 Cos(0.608)-1 3+sin(0.608)

= 06080000898

x_3 = 0607102

Hence, the real root is 0607102 Continuing further. we get the values near to it..