Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{131 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{131 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 6}{- \frac{393 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{131 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 6}{- \frac{393 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 3.5752428373$ $LaTeX: x_{2} = (3.5752428373) - \frac{- \frac{131 (3.5752428373)^{3}}{1000} + \sin{\left((3.5752428373) \right)} + 6}{- \frac{393 (3.5752428373)^{2}}{1000} + \cos{\left((3.5752428373) \right)}} = 3.5066374822$ $LaTeX: x_{3} = (3.5066374822) - \frac{- \frac{131 (3.5066374822)^{3}}{1000} + \sin{\left((3.5066374822) \right)} + 6}{- \frac{393 (3.5066374822)^{2}}{1000} + \cos{\left((3.5066374822) \right)}} = 3.5056609498$ $LaTeX: x_{4} = (3.5056609498) - \frac{- \frac{131 (3.5056609498)^{3}}{1000} + \sin{\left((3.5056609498) \right)} + 6}{- \frac{393 (3.5056609498)^{2}}{1000} + \cos{\left((3.5056609498) \right)}} = 3.5056607513$ $LaTeX: x_{5} = (3.5056607513) - \frac{- \frac{131 (3.5056607513)^{3}}{1000} + \sin{\left((3.5056607513) \right)} + 6}{- \frac{393 (3.5056607513)^{2}}{1000} + \cos{\left((3.5056607513) \right)}} = 3.5056607513$