Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{753 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{753 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 7}{- \frac{2259 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{753 (3.0000000000)^{3}}{1000} + \sin{\left((3.0000000000) \right)} + 7}{- \frac{2259 (3.0000000000)^{2}}{1000} + \cos{\left((3.0000000000) \right)}} = 2.3813665103$ $LaTeX: x_{2} = (2.3813665103) - \frac{- \frac{753 (2.3813665103)^{3}}{1000} + \sin{\left((2.3813665103) \right)} + 7}{- \frac{2259 (2.3813665103)^{2}}{1000} + \cos{\left((2.3813665103) \right)}} = 2.1981555042$ $LaTeX: x_{3} = (2.1981555042) - \frac{- \frac{753 (2.1981555042)^{3}}{1000} + \sin{\left((2.1981555042) \right)} + 7}{- \frac{2259 (2.1981555042)^{2}}{1000} + \cos{\left((2.1981555042) \right)}} = 2.1817923060$ $LaTeX: x_{4} = (2.1817923060) - \frac{- \frac{753 (2.1817923060)^{3}}{1000} + \sin{\left((2.1817923060) \right)} + 7}{- \frac{2259 (2.1817923060)^{2}}{1000} + \cos{\left((2.1817923060) \right)}} = 2.1816656106$ $LaTeX: x_{5} = (2.1816656106) - \frac{- \frac{753 (2.1816656106)^{3}}{1000} + \sin{\left((2.1816656106) \right)} + 7}{- \frac{2259 (2.1816656106)^{2}}{1000} + \cos{\left((2.1816656106) \right)}} = 2.1816656030$