Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{269 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{269 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 6}{- \frac{807 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{269 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 6}{- \frac{807 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.8640638541$ $LaTeX: x_{2} = (2.8640638541) - \frac{- \frac{269 (2.8640638541)^{3}}{1000} + \sin{\left((2.8640638541) \right)} + 6}{- \frac{807 (2.8640638541)^{2}}{1000} + \cos{\left((2.8640638541) \right)}} = 2.8580258232$ $LaTeX: x_{3} = (2.8580258232) - \frac{- \frac{269 (2.8580258232)^{3}}{1000} + \sin{\left((2.8580258232) \right)} + 6}{- \frac{807 (2.8580258232)^{2}}{1000} + \cos{\left((2.8580258232) \right)}} = 2.8580140070$ $LaTeX: x_{4} = (2.8580140070) - \frac{- \frac{269 (2.8580140070)^{3}}{1000} + \sin{\left((2.8580140070) \right)} + 6}{- \frac{807 (2.8580140070)^{2}}{1000} + \cos{\left((2.8580140070) \right)}} = 2.8580140069$ $LaTeX: x_{5} = (2.8580140069) - \frac{- \frac{269 (2.8580140069)^{3}}{1000} + \sin{\left((2.8580140069) \right)} + 6}{- \frac{807 (2.8580140069)^{2}}{1000} + \cos{\left((2.8580140069) \right)}} = 2.8580140069$