Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{21 x^{3}}{100} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{21 x_{n}^{3}}{100} + \sin{\left(x_{n} \right)} + 7}{- \frac{63 x_{n}^{2}}{100} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{21 (3.0000000000)^{3}}{100} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{63 (3.0000000000)^{2}}{100} + \cos{\left((3.0000000000) \right)}} = 3.2208891390$ $LaTeX: x_{2} = (3.2208891390) - \frac{- \frac{21 (3.2208891390)^{3}}{100} + \sin{\left((3.2208891390) \right)} + 7}{- \frac{63 (3.2208891390)^{2}}{100} + \cos{\left((3.2208891390) \right)}} = 3.2081265374$ $LaTeX: x_{3} = (3.2081265374) - \frac{- \frac{21 (3.2081265374)^{3}}{100} + \sin{\left((3.2081265374) \right)} + 7}{- \frac{63 (3.2081265374)^{2}}{100} + \cos{\left((3.2081265374) \right)}} = 3.2080832356$ $LaTeX: x_{4} = (3.2080832356) - \frac{- \frac{21 (3.2080832356)^{3}}{100} + \sin{\left((3.2080832356) \right)} + 7}{- \frac{63 (3.2080832356)^{2}}{100} + \cos{\left((3.2080832356) \right)}} = 3.2080832351$ $LaTeX: x_{5} = (3.2080832351) - \frac{- \frac{21 (3.2080832351)^{3}}{100} + \sin{\left((3.2080832351) \right)} + 7}{- \frac{63 (3.2080832351)^{2}}{100} + \cos{\left((3.2080832351) \right)}} = 3.2080832351$