Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{441 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{441 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 5}{- \frac{1323 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{441 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 5}{- \frac{1323 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.4753908717$ $LaTeX: x_{2} = (2.4753908717) - \frac{- \frac{441 (2.4753908717)^{3}}{1000} + \sin{\left((2.4753908717) \right)} + 5}{- \frac{1323 (2.4753908717)^{2}}{1000} + \cos{\left((2.4753908717) \right)}} = 2.3549435523$ $LaTeX: x_{3} = (2.3549435523) - \frac{- \frac{441 (2.3549435523)^{3}}{1000} + \sin{\left((2.3549435523) \right)} + 5}{- \frac{1323 (2.3549435523)^{2}}{1000} + \cos{\left((2.3549435523) \right)}} = 2.3485472597$ $LaTeX: x_{4} = (2.3485472597) - \frac{- \frac{441 (2.3485472597)^{3}}{1000} + \sin{\left((2.3485472597) \right)} + 5}{- \frac{1323 (2.3485472597)^{2}}{1000} + \cos{\left((2.3485472597) \right)}} = 2.3485295242$ $LaTeX: x_{5} = (2.3485295242) - \frac{- \frac{441 (2.3485295242)^{3}}{1000} + \sin{\left((2.3485295242) \right)} + 5}{- \frac{1323 (2.3485295242)^{2}}{1000} + \cos{\left((2.3485295242) \right)}} = 2.3485295241$