Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{23 x^{3}}{125} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{23 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 7}{- \frac{69 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{23 (3.0000000000)^{3}}{125} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{69 (3.0000000000)^{2}}{125} + \cos{\left((3.0000000000) \right)}} = 3.3647403063$ $LaTeX: x_{2} = (3.3647403063) - \frac{- \frac{23 (3.3647403063)^{3}}{125} + \sin{\left((3.3647403063) \right)} + 7}{- \frac{69 (3.3647403063)^{2}}{125} + \cos{\left((3.3647403063) \right)}} = 3.3328266651$ $LaTeX: x_{3} = (3.3328266651) - \frac{- \frac{23 (3.3328266651)^{3}}{125} + \sin{\left((3.3328266651) \right)} + 7}{- \frac{69 (3.3328266651)^{2}}{125} + \cos{\left((3.3328266651) \right)}} = 3.3325766697$ $LaTeX: x_{4} = (3.3325766697) - \frac{- \frac{23 (3.3325766697)^{3}}{125} + \sin{\left((3.3325766697) \right)} + 7}{- \frac{69 (3.3325766697)^{2}}{125} + \cos{\left((3.3325766697) \right)}} = 3.3325766543$ $LaTeX: x_{5} = (3.3325766543) - \frac{- \frac{23 (3.3325766543)^{3}}{125} + \sin{\left((3.3325766543) \right)} + 7}{- \frac{69 (3.3325766543)^{2}}{125} + \cos{\left((3.3325766543) \right)}} = 3.3325766543$