Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \sin{\left(x \right)}= \frac{741 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{741 x_{n}^{3}}{1000} + \sin{\left(x_{n} \right)} + 5}{- \frac{2223 x_{n}^{2}}{1000} + \cos{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{741 (1.0000000000)^{3}}{1000} + \sin{\left((1.0000000000) \right)} + 5}{- \frac{2223 (1.0000000000)^{2}}{1000} + \cos{\left((1.0000000000) \right)}} = 4.0311273395$ $LaTeX: x_{2} = (4.0311273395) - \frac{- \frac{741 (4.0311273395)^{3}}{1000} + \sin{\left((4.0311273395) \right)} + 5}{- \frac{2223 (4.0311273395)^{2}}{1000} + \cos{\left((4.0311273395) \right)}} = 2.8253494221$ $LaTeX: x_{3} = (2.8253494221) - \frac{- \frac{741 (2.8253494221)^{3}}{1000} + \sin{\left((2.8253494221) \right)} + 5}{- \frac{2223 (2.8253494221)^{2}}{1000} + \cos{\left((2.8253494221) \right)}} = 2.2155179749$ $LaTeX: x_{4} = (2.2155179749) - \frac{- \frac{741 (2.2155179749)^{3}}{1000} + \sin{\left((2.2155179749) \right)} + 5}{- \frac{2223 (2.2155179749)^{2}}{1000} + \cos{\left((2.2155179749) \right)}} = 2.0192946061$ $LaTeX: x_{5} = (2.0192946061) - \frac{- \frac{741 (2.0192946061)^{3}}{1000} + \sin{\left((2.0192946061) \right)} + 5}{- \frac{2223 (2.0192946061)^{2}}{1000} + \cos{\left((2.0192946061) \right)}} = 1.9982239942$