Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{537 x^{3}}{1000} - 8$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{537 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 8}{- \frac{1611 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{537 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 8}{- \frac{1611 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.5895226889$ $LaTeX: x_{2} = (2.5895226889) - \frac{- \frac{537 (2.5895226889)^{3}}{1000} + \sin{\left((2.5895226889) \right)} + 8}{- \frac{1611 (2.5895226889)^{2}}{1000} + \cos{\left((2.5895226889) \right)}} = 2.5208591853$ $LaTeX: x_{3} = (2.5208591853) - \frac{- \frac{537 (2.5208591853)^{3}}{1000} + \sin{\left((2.5208591853) \right)} + 8}{- \frac{1611 (2.5208591853)^{2}}{1000} + \cos{\left((2.5208591853) \right)}} = 2.5189791411$ $LaTeX: x_{4} = (2.5189791411) - \frac{- \frac{537 (2.5189791411)^{3}}{1000} + \sin{\left((2.5189791411) \right)} + 8}{- \frac{1611 (2.5189791411)^{2}}{1000} + \cos{\left((2.5189791411) \right)}} = 2.5189777473$ $LaTeX: x_{5} = (2.5189777473) - \frac{- \frac{537 (2.5189777473)^{3}}{1000} + \sin{\left((2.5189777473) \right)} + 8}{- \frac{1611 (2.5189777473)^{2}}{1000} + \cos{\left((2.5189777473) \right)}} = 2.5189777473$