use the iterative method to find a real root of the following equations correct to four sig. fig. \( \begin{array}{ll}\text { (1) } \cos x=3 x-1 & \text { (2) } \sin x=10(x-1) \\ \text { (3) } e^{x}=\cot x & \text { (4) } 1+x^{2}=x^{3} \\ \text { (5) } \sin ^{2} x=x^{2}-1 & \text { (6) } x=(5-x)^{1 / 3}\end{array} \)
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To find a real root of the given equations using the iterative method, we can rearrange each equation into the form \( x = g(x) \) and then apply the iteration. Let's start with the steps for each equation. 1. **For \( \cos x = 3x - 1 \)**: Rearranged as: \[ x = \frac{\cos x + 1}{3} \] Using initial guess \( x_0 = 0 \): \[ x_{1} = \frac{\cos(0) + 1}{3} = \frac{1 + 1}{3} = \frac{2}{3} \approx 0.6667 \] Continuing iterations will converge to the root. 2. **For \( \sin x = 10(x-1) \)**: Rearranged as: \[ x = 1 + \frac{\sin x}{10} \] Using initial guess \( x_0 = 1 \): \[ x_{1} = 1 + \frac{\sin(1)}{10} \approx 1 + 0.0840 \approx 1.0840 \] Repeating this process approaches the root. 3. **For \( e^{x} = \cot x \)**: Rearranged as: \[ x = \cot^{-1}(e^{x}) \] A good initial guess would be \( x_0 = 0 \): \[ x_{1} = \cot^{-1}(e^{0}) = \cot^{-1}(1) \approx 0.7854 \] More iterations yield the required precision. 4. **For \( 1 + x^{2} = x^{3} \)**: Rearranged as: \[ x = 1 + \frac{x^{2}}{x^{2}} \] Starting with \( x_0 = 1 \): \[ x_{1} = 1 + \frac{1}{1} = 2 \] Iterating further will refine this guess. 5. **For \( \sin^{2} x = x^{2} - 1 \)**: Rearranged as: \[ x = \sqrt{\sin^{2} x + 1} \] Start with \( x_0 = 1 \): \[ x_{1} = \sqrt{\sin^{2}(1) + 1} \approx \sqrt{0.708 + 1} \approx 1.2910 \] Keep iterating for higher precision. 6. **For \( x = (5 - x)^{1/3} \)**: Rearranged as: \[ x = (5 - x)^{1/3} \] Begin with \( x_0 = 1 \): \[ x_{1} = (5 - 1)^{1/3} \approx 1.5874 \] Further iterations will approach the real root. After iterating through a few additional steps for each equation, round the result to four significant figures. Continue this iterative process until the results stabilize.