Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{113 x^{3}}{500} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{113 x_{n}^{3}}{500} + \sin{\left(x_{n} \right)} + 7}{- \frac{339 x_{n}^{2}}{500} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{113 (3.0000000000)^{3}}{500} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{339 (3.0000000000)^{2}}{500} + \cos{\left((3.0000000000) \right)}} = 3.1465201787$ $LaTeX: x_{2} = (3.1465201787) - \frac{- \frac{113 (3.1465201787)^{3}}{500} + \sin{\left((3.1465201787) \right)} + 7}{- \frac{339 (3.1465201787)^{2}}{500} + \cos{\left((3.1465201787) \right)}} = 3.1406374880$ $LaTeX: x_{3} = (3.1406374880) - \frac{- \frac{113 (3.1406374880)^{3}}{500} + \sin{\left((3.1406374880) \right)} + 7}{- \frac{339 (3.1406374880)^{2}}{500} + \cos{\left((3.1406374880) \right)}} = 3.1406278972$ $LaTeX: x_{4} = (3.1406278972) - \frac{- \frac{113 (3.1406278972)^{3}}{500} + \sin{\left((3.1406278972) \right)} + 7}{- \frac{339 (3.1406278972)^{2}}{500} + \cos{\left((3.1406278972) \right)}} = 3.1406278972$ $LaTeX: x_{5} = (3.1406278972) - \frac{- \frac{113 (3.1406278972)^{3}}{500} + \sin{\left((3.1406278972) \right)} + 7}{- \frac{339 (3.1406278972)^{2}}{500} + \cos{\left((3.1406278972) \right)}} = 3.1406278972$