Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{537 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{537 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 6}{- \frac{1611 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{537 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 6}{- \frac{1611 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.4603987320$ $LaTeX: x_{2} = (2.4603987320) - \frac{- \frac{537 (2.4603987320)^{3}}{1000} + \sin{\left((2.4603987320) \right)} + 6}{- \frac{1611 (2.4603987320)^{2}}{1000} + \cos{\left((2.4603987320) \right)}} = 2.3304303249$ $LaTeX: x_{3} = (2.3304303249) - \frac{- \frac{537 (2.3304303249)^{3}}{1000} + \sin{\left((2.3304303249) \right)} + 6}{- \frac{1611 (2.3304303249)^{2}}{1000} + \cos{\left((2.3304303249) \right)}} = 2.3228682003$ $LaTeX: x_{4} = (2.3228682003) - \frac{- \frac{537 (2.3228682003)^{3}}{1000} + \sin{\left((2.3228682003) \right)} + 6}{- \frac{1611 (2.3228682003)^{2}}{1000} + \cos{\left((2.3228682003) \right)}} = 2.3228431094$ $LaTeX: x_{5} = (2.3228431094) - \frac{- \frac{537 (2.3228431094)^{3}}{1000} + \sin{\left((2.3228431094) \right)} + 6}{- \frac{1611 (2.3228431094)^{2}}{1000} + \cos{\left((2.3228431094) \right)}} = 2.3228431091$