Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{49 x^{3}}{100} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{49 x_{n}^{3}}{100} + \sin{\left(x_{n} \right)} + 1}{- \frac{147 x_{n}^{2}}{100} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{49 (1.0000000000)^{3}}{100} + \sin{\left((1.0000000000) \right)} + 1}{- \frac{147 (1.0000000000)^{2}}{100} + \cos{\left((1.0000000000) \right)}} = 2.4536671365$ $LaTeX: x_{2} = (2.4536671365) - \frac{- \frac{49 (2.4536671365)^{3}}{100} + \sin{\left((2.4536671365) \right)} + 1}{- \frac{147 (2.4536671365)^{2}}{100} + \cos{\left((2.4536671365) \right)}} = 1.8713475593$ $LaTeX: x_{3} = (1.8713475593) - \frac{- \frac{49 (1.8713475593)^{3}}{100} + \sin{\left((1.8713475593) \right)} + 1}{- \frac{147 (1.8713475593)^{2}}{100} + \cos{\left((1.8713475593) \right)}} = 1.6406365199$ $LaTeX: x_{4} = (1.6406365199) - \frac{- \frac{49 (1.6406365199)^{3}}{100} + \sin{\left((1.6406365199) \right)} + 1}{- \frac{147 (1.6406365199)^{2}}{100} + \cos{\left((1.6406365199) \right)}} = 1.5993313415$ $LaTeX: x_{5} = (1.5993313415) - \frac{- \frac{49 (1.5993313415)^{3}}{100} + \sin{\left((1.5993313415) \right)} + 1}{- \frac{147 (1.5993313415)^{2}}{100} + \cos{\left((1.5993313415) \right)}} = 1.5980295758$