Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{x^{3}}{4} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{4} + \sin{\left(x_{n} \right)} + 7}{- \frac{3 x_{n}^{2}}{4} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{(3.0000000000)^{3}}{4} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{3 (3.0000000000)^{2}}{4} + \cos{\left((3.0000000000) \right)}} = 3.0505323498$ $LaTeX: x_{2} = (3.0505323498) - \frac{- \frac{(3.0505323498)^{3}}{4} + \sin{\left((3.0505323498) \right)} + 7}{- \frac{3 (3.0505323498)^{2}}{4} + \cos{\left((3.0505323498) \right)}} = 3.0497879735$ $LaTeX: x_{3} = (3.0497879735) - \frac{- \frac{(3.0497879735)^{3}}{4} + \sin{\left((3.0497879735) \right)} + 7}{- \frac{3 (3.0497879735)^{2}}{4} + \cos{\left((3.0497879735) \right)}} = 3.0497878114$ $LaTeX: x_{4} = (3.0497878114) - \frac{- \frac{(3.0497878114)^{3}}{4} + \sin{\left((3.0497878114) \right)} + 7}{- \frac{3 (3.0497878114)^{2}}{4} + \cos{\left((3.0497878114) \right)}} = 3.0497878114$ $LaTeX: x_{5} = (3.0497878114) - \frac{- \frac{(3.0497878114)^{3}}{4} + \sin{\left((3.0497878114) \right)} + 7}{- \frac{3 (3.0497878114)^{2}}{4} + \cos{\left((3.0497878114) \right)}} = 3.0497878114$