To find the positive root of the equation $2x = 3 + \cos(x)$ using the Bisection Method, follow the steps below:

Steps:

1. Rewrite the equation in the form $f(x) = 0$: $f(x) = 2x - 3 - \cos(x)$ The root of the equation corresponds to the value of $x$ where $f(x) = 0$.

2. Choose an initial interval $[a, b]$ such that $f(a) \times f(b) < 0$. This ensures that there is a root in the interval.

3. Apply the Bisection Method formula to find the midpoint $c$: $c = \frac{a + b}{2}$ Calculate $f(c)$ and check the sign of $f(c)$.

4. Update the interval:

   • If $f(a) \times f(c) < 0$, then the root lies in the interval $[a, c]$, so update $b = c$.
   • If $f(b) \times f(c) < 0$, then the root lies in the interval $[c, b]$, so update $a = c$.

5. Repeat the process until the interval is sufficiently small (i.e., until $|b - a|$ is smaller than the desired tolerance).

Step-by-Step Calculation:

1. Initial Interval: Let's find an interval $[a, b]$ where $f(x)$ changes sign.

   • For $x = 0$: $f(0) = 2(0) - 3 - \cos(0) = -3 - 1 = -4$
   • For $x = 2$: $f(2) = 2(2) - 3 - \cos(2) = 4 - 3 - (-0.416) = 1 + 0.416 \approx 1.416$

   Since $f(0) = -4$ and $f(2) = 1.416$, the function changes sign between $x = 0$ and $x = 2$. Therefore, the interval $[0, 2]$ is a suitable starting point.

2. First Iteration:

   • $a = 0$, $b = 2$
   • Midpoint $c = \frac{0 + 2}{2} = 1$
   • $f(1)$: $f(1) = 2(1) - 3 - \cos(1) = 2 - 3 - 0.540 = -1.540$ Since $f(1) < 0$, the root lies in the interval $[1, 2]$. Update $a = 1$.

3. Second Iteration:

   • $a = 1$, $b = 2$
   • Midpoint $c = \frac{1 + 2}{2} = 1.5$
   • $f(1.5)$: $f(1.5) = 2(1.5) - 3 - \cos(1.5) = 3 - 3 - 0.0707 = -0.0707$ Since $f(1.5) < 0$, the root lies in the interval $[1.5, 2]$. Update $a = 1.5$.

4. Third Iteration:

   • $a = 1.5$, $b = 2$
   • Midpoint $c = \frac{1.5 + 2}{2} = 1.75$
   • $f(1.75)$: $f(1.75) = 2(1.75) - 3 - \cos(1.75) = 3.5 - 3 - (-0.178) = 0.678$ Since $f(1.75) > 0$, the root lies in the interval $[1.5, 1.75]$. Update $b = 1.75$.

5. Repeat the process until the interval is sufficiently small. Continue bisecting until the root is accurate to the desired number of decimal places.

Final Answer:

The Bisection Method will give an increasingly accurate approximation of the root within the interval $[1.5, 1.75]$. After several iterations, you'll converge on the root.

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Would you like me to complete more iterations, or explain anything further?

5 Related Questions:

1. What are the advantages of the Bisection Method over other numerical methods?
2. How does the choice of initial interval affect the convergence of the Bisection Method?
3. Can the Bisection Method be applied to non-continuous functions?
4. How do you determine the tolerance level for the Bisection Method?
5. What is the convergence rate of the Bisection Method compared to Newton's Method?

Tip: Always ensure that the function changes sign within the interval before applying the Bisection Method.