Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{483 x^{3}}{500} - 7$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{483 x_{n}^{3}}{500} + \cos{\left(x_{n} \right)} + 7}{- \frac{1449 x_{n}^{2}}{500} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{483 (1.0000000000)^{3}}{500} + \cos{\left((1.0000000000) \right)} + 7}{- \frac{1449 (1.0000000000)^{2}}{500} - \sin{\left((1.0000000000) \right)}} = 2.7580835184$ $LaTeX: x_{2} = (2.7580835184) - \frac{- \frac{483 (2.7580835184)^{3}}{500} + \cos{\left((2.7580835184) \right)} + 7}{- \frac{1449 (2.7580835184)^{2}}{500} - \sin{\left((2.7580835184) \right)}} = 2.1249327355$ $LaTeX: x_{3} = (2.1249327355) - \frac{- \frac{483 (2.1249327355)^{3}}{500} + \cos{\left((2.1249327355) \right)} + 7}{- \frac{1449 (2.1249327355)^{2}}{500} - \sin{\left((2.1249327355) \right)}} = 1.9243861906$ $LaTeX: x_{4} = (1.9243861906) - \frac{- \frac{483 (1.9243861906)^{3}}{500} + \cos{\left((1.9243861906) \right)} + 7}{- \frac{1449 (1.9243861906)^{2}}{500} - \sin{\left((1.9243861906) \right)}} = 1.9046373047$ $LaTeX: x_{5} = (1.9046373047) - \frac{- \frac{483 (1.9046373047)^{3}}{500} + \cos{\left((1.9046373047) \right)} + 7}{- \frac{1449 (1.9046373047)^{2}}{500} - \sin{\left((1.9046373047) \right)}} = 1.9044539064$