Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{4 x^{3}}{25} - 2$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{4 x_{n}^{3}}{25} + \sin{\left(x_{n} \right)} + 2}{- \frac{12 x_{n}^{2}}{25} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{4 (3.0000000000)^{3}}{25} + \sin{\left((3.0000000000) \right)} + 2}{- \frac{12 (3.0000000000)^{2}}{25} + \cos{\left((3.0000000000) \right)}} = 2.5896642051$ $LaTeX: x_{2} = (2.5896642051) - \frac{- \frac{4 (2.5896642051)^{3}}{25} + \sin{\left((2.5896642051) \right)} + 2}{- \frac{12 (2.5896642051)^{2}}{25} + \cos{\left((2.5896642051) \right)}} = 2.5271605824$ $LaTeX: x_{3} = (2.5271605824) - \frac{- \frac{4 (2.5271605824)^{3}}{25} + \sin{\left((2.5271605824) \right)} + 2}{- \frac{12 (2.5271605824)^{2}}{25} + \cos{\left((2.5271605824) \right)}} = 2.5256472747$ $LaTeX: x_{4} = (2.5256472747) - \frac{- \frac{4 (2.5256472747)^{3}}{25} + \sin{\left((2.5256472747) \right)} + 2}{- \frac{12 (2.5256472747)^{2}}{25} + \cos{\left((2.5256472747) \right)}} = 2.5256463882$ $LaTeX: x_{5} = (2.5256463882) - \frac{- \frac{4 (2.5256463882)^{3}}{25} + \sin{\left((2.5256463882) \right)} + 2}{- \frac{12 (2.5256463882)^{2}}{25} + \cos{\left((2.5256463882) \right)}} = 2.5256463882$