Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{111 x^{3}}{1000} - 9$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{111 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 9}{- \frac{333 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{111 (5.0000000000)^{3}}{1000} + \sin{\left((5.0000000000) \right)} + 9}{- \frac{333 (5.0000000000)^{2}}{1000} + \cos{\left((5.0000000000) \right)}} = 4.2745082461$ $LaTeX: x_{2} = (4.2745082461) - \frac{- \frac{111 (4.2745082461)^{3}}{1000} + \sin{\left((4.2745082461) \right)} + 9}{- \frac{333 (4.2745082461)^{2}}{1000} + \cos{\left((4.2745082461) \right)}} = 4.1861761633$ $LaTeX: x_{3} = (4.1861761633) - \frac{- \frac{111 (4.1861761633)^{3}}{1000} + \sin{\left((4.1861761633) \right)} + 9}{- \frac{333 (4.1861761633)^{2}}{1000} + \cos{\left((4.1861761633) \right)}} = 4.1849852842$ $LaTeX: x_{4} = (4.1849852842) - \frac{- \frac{111 (4.1849852842)^{3}}{1000} + \sin{\left((4.1849852842) \right)} + 9}{- \frac{333 (4.1849852842)^{2}}{1000} + \cos{\left((4.1849852842) \right)}} = 4.1849850689$ $LaTeX: x_{5} = (4.1849850689) - \frac{- \frac{111 (4.1849850689)^{3}}{1000} + \sin{\left((4.1849850689) \right)} + 9}{- \frac{333 (4.1849850689)^{2}}{1000} + \cos{\left((4.1849850689) \right)}} = 4.1849850689$