Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{48 x^{3}}{125} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{48 x_{n}^{3}}{125} + \sin{\left(x_{n} \right)} + 6}{- \frac{144 x_{n}^{2}}{125} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{48 (3.0000000000)^{3}}{125} + \sin{\left((3.0000000000) \right)} + 6}{- \frac{144 (3.0000000000)^{2}}{125} + \cos{\left((3.0000000000) \right)}} = 2.6278497284$ $LaTeX: x_{2} = (2.6278497284) - \frac{- \frac{48 (2.6278497284)^{3}}{125} + \sin{\left((2.6278497284) \right)} + 6}{- \frac{144 (2.6278497284)^{2}}{125} + \cos{\left((2.6278497284) \right)}} = 2.5738108818$ $LaTeX: x_{3} = (2.5738108818) - \frac{- \frac{48 (2.5738108818)^{3}}{125} + \sin{\left((2.5738108818) \right)} + 6}{- \frac{144 (2.5738108818)^{2}}{125} + \cos{\left((2.5738108818) \right)}} = 2.5726875213$ $LaTeX: x_{4} = (2.5726875213) - \frac{- \frac{48 (2.5726875213)^{3}}{125} + \sin{\left((2.5726875213) \right)} + 6}{- \frac{144 (2.5726875213)^{2}}{125} + \cos{\left((2.5726875213) \right)}} = 2.5726870394$ $LaTeX: x_{5} = (2.5726870394) - \frac{- \frac{48 (2.5726870394)^{3}}{125} + \sin{\left((2.5726870394) \right)} + 6}{- \frac{144 (2.5726870394)^{2}}{125} + \cos{\left((2.5726870394) \right)}} = 2.5726870394$