Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{363 x^{3}}{500} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{363 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 7}{- \frac{1089 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{363 (3.0000000000)^{3}}{500} + \cos{\left((3.0000000000) \right)} + 7}{- \frac{1089 (3.0000000000)^{2}}{500} - \sin{\left((3.0000000000) \right)}} = 2.3115580267$ $LaTeX: x_{2} = (2.3115580267) - \frac{- \frac{363 (2.3115580267)^{3}}{500} + \cos{\left((2.3115580267) \right)} + 7}{- \frac{1089 (2.3115580267)^{2}}{500} - \sin{\left((2.3115580267) \right)}} = 2.0980801983$ $LaTeX: x_{3} = (2.0980801983) - \frac{- \frac{363 (2.0980801983)^{3}}{500} + \cos{\left((2.0980801983) \right)} + 7}{- \frac{1089 (2.0980801983)^{2}}{500} - \sin{\left((2.0980801983) \right)}} = 2.0781549130$ $LaTeX: x_{4} = (2.0781549130) - \frac{- \frac{363 (2.0781549130)^{3}}{500} + \cos{\left((2.0781549130) \right)} + 7}{- \frac{1089 (2.0781549130)^{2}}{500} - \sin{\left((2.0781549130) \right)}} = 2.0779886006$ $LaTeX: x_{5} = (2.0779886006) - \frac{- \frac{363 (2.0779886006)^{3}}{500} + \cos{\left((2.0779886006) \right)} + 7}{- \frac{1089 (2.0779886006)^{2}}{500} - \sin{\left((2.0779886006) \right)}} = 2.0779885891$