Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{59 x^{3}}{500} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{59 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 7}{- \frac{177 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{59 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{177 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 3.9471089834$ $LaTeX: x_{2} = (3.9471089834) - \frac{- \frac{59 (3.9471089834)^{3}}{500} + \sin{\left((3.9471089834) \right)} + 7}{- \frac{177 (3.9471089834)^{2}}{500} + \cos{\left((3.9471089834) \right)}} = 3.7896401375$ $LaTeX: x_{3} = (3.7896401375) - \frac{- \frac{59 (3.7896401375)^{3}}{500} + \sin{\left((3.7896401375) \right)} + 7}{- \frac{177 (3.7896401375)^{2}}{500} + \cos{\left((3.7896401375) \right)}} = 3.7852678928$ $LaTeX: x_{4} = (3.7852678928) - \frac{- \frac{59 (3.7852678928)^{3}}{500} + \sin{\left((3.7852678928) \right)} + 7}{- \frac{177 (3.7852678928)^{2}}{500} + \cos{\left((3.7852678928) \right)}} = 3.7852645078$ $LaTeX: x_{5} = (3.7852645078) - \frac{- \frac{59 (3.7852645078)^{3}}{500} + \sin{\left((3.7852645078) \right)} + 7}{- \frac{177 (3.7852645078)^{2}}{500} + \cos{\left((3.7852645078) \right)}} = 3.7852645078$