Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{523 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{523 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 5}{- \frac{1569 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{523 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{1569 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.4057385712$ $LaTeX: x_{2} = (2.4057385712) - \frac{- \frac{523 (2.4057385712)^{3}}{1000} + \sin{\left((2.4057385712) \right)} + 5}{- \frac{1569 (2.4057385712)^{2}}{1000} + \cos{\left((2.4057385712) \right)}} = 2.2417472363$ $LaTeX: x_{3} = (2.2417472363) - \frac{- \frac{523 (2.2417472363)^{3}}{1000} + \sin{\left((2.2417472363) \right)} + 5}{- \frac{1569 (2.2417472363)^{2}}{1000} + \cos{\left((2.2417472363) \right)}} = 2.2289626237$ $LaTeX: x_{4} = (2.2289626237) - \frac{- \frac{523 (2.2289626237)^{3}}{1000} + \sin{\left((2.2289626237) \right)} + 5}{- \frac{1569 (2.2289626237)^{2}}{1000} + \cos{\left((2.2289626237) \right)}} = 2.2288867311$ $LaTeX: x_{5} = (2.2288867311) - \frac{- \frac{523 (2.2288867311)^{3}}{1000} + \sin{\left((2.2288867311) \right)} + 5}{- \frac{1569 (2.2288867311)^{2}}{1000} + \cos{\left((2.2288867311) \right)}} = 2.2288867284$