Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{433 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{433 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 5}{- \frac{1299 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{433 (1.0000000000)^{3}}{500} + \sin{\left((1.0000000000) \right)} + 5}{- \frac{1299 (1.0000000000)^{2}}{500} + \cos{\left((1.0000000000) \right)}} = 3.4179795696$ $LaTeX: x_{2} = (3.4179795696) - \frac{- \frac{433 (3.4179795696)^{3}}{500} + \sin{\left((3.4179795696) \right)} + 5}{- \frac{1299 (3.4179795696)^{2}}{500} + \cos{\left((3.4179795696) \right)}} = 2.4646183075$ $LaTeX: x_{3} = (2.4646183075) - \frac{- \frac{433 (2.4646183075)^{3}}{500} + \sin{\left((2.4646183075) \right)} + 5}{- \frac{1299 (2.4646183075)^{2}}{500} + \cos{\left((2.4646183075) \right)}} = 2.0214950045$ $LaTeX: x_{4} = (2.0214950045) - \frac{- \frac{433 (2.0214950045)^{3}}{500} + \sin{\left((2.0214950045) \right)} + 5}{- \frac{1299 (2.0214950045)^{2}}{500} + \cos{\left((2.0214950045) \right)}} = 1.9080654855$ $LaTeX: x_{5} = (1.9080654855) - \frac{- \frac{433 (1.9080654855)^{3}}{500} + \sin{\left((1.9080654855) \right)} + 5}{- \frac{1299 (1.9080654855)^{2}}{500} + \cos{\left((1.9080654855) \right)}} = 1.9006904171$