Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{289 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{289 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 7}{- \frac{867 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{289 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{867 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.9247264237$ $LaTeX: x_{2} = (2.9247264237) - \frac{- \frac{289 (2.9247264237)^{3}}{1000} + \sin{\left((2.9247264237) \right)} + 7}{- \frac{867 (2.9247264237)^{2}}{1000} + \cos{\left((2.9247264237) \right)}} = 2.9229291627$ $LaTeX: x_{3} = (2.9229291627) - \frac{- \frac{289 (2.9229291627)^{3}}{1000} + \sin{\left((2.9229291627) \right)} + 7}{- \frac{867 (2.9229291627)^{2}}{1000} + \cos{\left((2.9229291627) \right)}} = 2.9229281443$ $LaTeX: x_{4} = (2.9229281443) - \frac{- \frac{289 (2.9229281443)^{3}}{1000} + \sin{\left((2.9229281443) \right)} + 7}{- \frac{867 (2.9229281443)^{2}}{1000} + \cos{\left((2.9229281443) \right)}} = 2.9229281443$ $LaTeX: x_{5} = (2.9229281443) - \frac{- \frac{289 (2.9229281443)^{3}}{1000} + \sin{\left((2.9229281443) \right)} + 7}{- \frac{867 (2.9229281443)^{2}}{1000} + \cos{\left((2.9229281443) \right)}} = 2.9229281443$