Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{411 x^{3}}{500} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{411 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 4}{- \frac{1233 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{411 (1.0000000000)^{3}}{500} + \sin{\left((1.0000000000) \right)} + 4}{- \frac{1233 (1.0000000000)^{2}}{500} + \cos{\left((1.0000000000) \right)}} = 3.0872803644$ $LaTeX: x_{2} = (3.0872803644) - \frac{- \frac{411 (3.0872803644)^{3}}{500} + \sin{\left((3.0872803644) \right)} + 4}{- \frac{1233 (3.0872803644)^{2}}{500} + \cos{\left((3.0872803644) \right)}} = 2.2655868794$ $LaTeX: x_{3} = (2.2655868794) - \frac{- \frac{411 (2.2655868794)^{3}}{500} + \sin{\left((2.2655868794) \right)} + 4}{- \frac{1233 (2.2655868794)^{2}}{500} + \cos{\left((2.2655868794) \right)}} = 1.9053164795$ $LaTeX: x_{4} = (1.9053164795) - \frac{- \frac{411 (1.9053164795)^{3}}{500} + \sin{\left((1.9053164795) \right)} + 4}{- \frac{1233 (1.9053164795)^{2}}{500} + \cos{\left((1.9053164795) \right)}} = 1.8254722850$ $LaTeX: x_{5} = (1.8254722850) - \frac{- \frac{411 (1.8254722850)^{3}}{500} + \sin{\left((1.8254722850) \right)} + 4}{- \frac{1233 (1.8254722850)^{2}}{500} + \cos{\left((1.8254722850) \right)}} = 1.8216264706$