Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{369 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{369 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 5}{- \frac{1107 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{369 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{1107 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.5597659732$ $LaTeX: x_{2} = (2.5597659732) - \frac{- \frac{369 (2.5597659732)^{3}}{1000} + \sin{\left((2.5597659732) \right)} + 5}{- \frac{1107 (2.5597659732)^{2}}{1000} + \cos{\left((2.5597659732) \right)}} = 2.4807022433$ $LaTeX: x_{3} = (2.4807022433) - \frac{- \frac{369 (2.4807022433)^{3}}{1000} + \sin{\left((2.4807022433) \right)} + 5}{- \frac{1107 (2.4807022433)^{2}}{1000} + \cos{\left((2.4807022433) \right)}} = 2.4781611864$ $LaTeX: x_{4} = (2.4781611864) - \frac{- \frac{369 (2.4781611864)^{3}}{1000} + \sin{\left((2.4781611864) \right)} + 5}{- \frac{1107 (2.4781611864)^{2}}{1000} + \cos{\left((2.4781611864) \right)}} = 2.4781585883$ $LaTeX: x_{5} = (2.4781585883) - \frac{- \frac{369 (2.4781585883)^{3}}{1000} + \sin{\left((2.4781585883) \right)} + 5}{- \frac{1107 (2.4781585883)^{2}}{1000} + \cos{\left((2.4781585883) \right)}} = 2.4781585883$