Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{31 x^{3}}{200} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{31 x_{n}^{3}}{200} + \sin{\left(x_{n} \right)} + 4}{- \frac{93 x_{n}^{2}}{200} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{31 (3.0000000000)^{3}}{200} + \sin{\left((3.0000000000) \right)} + 4}{- \frac{93 (3.0000000000)^{2}}{200} + \cos{\left((3.0000000000) \right)}} = 2.9915207622$ $LaTeX: x_{2} = (2.9915207622) - \frac{- \frac{31 (2.9915207622)^{3}}{200} + \sin{\left((2.9915207622) \right)} + 4}{- \frac{93 (2.9915207622)^{2}}{200} + \cos{\left((2.9915207622) \right)}} = 2.9915003014$ $LaTeX: x_{3} = (2.9915003014) - \frac{- \frac{31 (2.9915003014)^{3}}{200} + \sin{\left((2.9915003014) \right)} + 4}{- \frac{93 (2.9915003014)^{2}}{200} + \cos{\left((2.9915003014) \right)}} = 2.9915003013$ $LaTeX: x_{4} = (2.9915003013) - \frac{- \frac{31 (2.9915003013)^{3}}{200} + \sin{\left((2.9915003013) \right)} + 4}{- \frac{93 (2.9915003013)^{2}}{200} + \cos{\left((2.9915003013) \right)}} = 2.9915003013$ $LaTeX: x_{5} = (2.9915003013) - \frac{- \frac{31 (2.9915003013)^{3}}{200} + \sin{\left((2.9915003013) \right)} + 4}{- \frac{93 (2.9915003013)^{2}}{200} + \cos{\left((2.9915003013) \right)}} = 2.9915003013$