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

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{3 x_{n}^{3}}{100} + \sin{\left(x_{n} \right)} + 3}{- \frac{9 x_{n}^{2}}{100} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{3 (5.0000000000)^{3}}{100} + \sin{\left((5.0000000000) \right)} + 3}{- \frac{9 (5.0000000000)^{2}}{100} + \cos{\left((5.0000000000) \right)}} = 4.1309101305$ $LaTeX: x_{2} = (4.1309101305) - \frac{- \frac{3 (4.1309101305)^{3}}{100} + \sin{\left((4.1309101305) \right)} + 3}{- \frac{9 (4.1309101305)^{2}}{100} + \cos{\left((4.1309101305) \right)}} = 4.1546990760$ $LaTeX: x_{3} = (4.1546990760) - \frac{- \frac{3 (4.1546990760)^{3}}{100} + \sin{\left((4.1546990760) \right)} + 3}{- \frac{9 (4.1546990760)^{2}}{100} + \cos{\left((4.1546990760) \right)}} = 4.1547119791$ $LaTeX: x_{4} = (4.1547119791) - \frac{- \frac{3 (4.1547119791)^{3}}{100} + \sin{\left((4.1547119791) \right)} + 3}{- \frac{9 (4.1547119791)^{2}}{100} + \cos{\left((4.1547119791) \right)}} = 4.1547119791$ $LaTeX: x_{5} = (4.1547119791) - \frac{- \frac{3 (4.1547119791)^{3}}{100} + \sin{\left((4.1547119791) \right)} + 3}{- \frac{9 (4.1547119791)^{2}}{100} + \cos{\left((4.1547119791) \right)}} = 4.1547119791$