Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{19 x^{3}}{40} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{19 x_{n}^{3}}{40} + \sin{\left(x_{n} \right)} + 7}{- \frac{57 x_{n}^{2}}{40} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{19 (3.0000000000)^{3}}{40} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{57 (3.0000000000)^{2}}{40} + \cos{\left((3.0000000000) \right)}} = 2.5885716193$ $LaTeX: x_{2} = (2.5885716193) - \frac{- \frac{19 (2.5885716193)^{3}}{40} + \sin{\left((2.5885716193) \right)} + 7}{- \frac{57 (2.5885716193)^{2}}{40} + \cos{\left((2.5885716193) \right)}} = 2.5199397484$ $LaTeX: x_{3} = (2.5199397484) - \frac{- \frac{19 (2.5199397484)^{3}}{40} + \sin{\left((2.5199397484) \right)} + 7}{- \frac{57 (2.5199397484)^{2}}{40} + \cos{\left((2.5199397484) \right)}} = 2.5180634237$ $LaTeX: x_{4} = (2.5180634237) - \frac{- \frac{19 (2.5180634237)^{3}}{40} + \sin{\left((2.5180634237) \right)} + 7}{- \frac{57 (2.5180634237)^{2}}{40} + \cos{\left((2.5180634237) \right)}} = 2.5180620360$ $LaTeX: x_{5} = (2.5180620360) - \frac{- \frac{19 (2.5180620360)^{3}}{40} + \sin{\left((2.5180620360) \right)} + 7}{- \frac{57 (2.5180620360)^{2}}{40} + \cos{\left((2.5180620360) \right)}} = 2.5180620360$