To solve $2x = 3 + \cos x$ using the bisection method, we first rearrange the equation into the form:

$f(x) = 2x - 3 - \cos x = 0$

We now use the bisection method to find the positive root of this equation. The bisection method involves the following steps:

Steps for the Bisection Method:

1. Select two initial points: Choose two values $a$ and $b$ such that $f(a)$ and $f(b)$ have opposite signs. This ensures that there is a root in the interval $[a, b]$.
2. Compute the midpoint: $c = \frac{a+b}{2}$ and evaluate $f(c)$.
3. Determine the new interval:
   • If $f(c) = 0$, then $c$ is the root.
   • If $f(a) \cdot f(c) < 0$, the root lies in the interval $[a, c]$, so set $b = c$.
   • If $f(c) \cdot f(b) < 0$, the root lies in the interval $[c, b]$, so set $a = c$.
4. Repeat the process until the desired accuracy is achieved.

Step 1: Select an interval

We start by choosing an interval where the sign of $f(x)$ changes. Let’s try initial guesses, $a = 1$ and $b = 2$.

• $f(1) = 2(1) - 3 - \cos(1) = 2 - 3 - 0.5403 = -1.5403$
• $f(2) = 2(2) - 3 - \cos(2) = 4 - 3 - (-0.4161) = 1.4161$

Since $f(1)$ is negative and $f(2)$ is positive, there is a root between $1$ and $2$.

Step 2: Apply the bisection method

We now begin the iterative process:

1. Midpoint $c_1 = \frac{1 + 2}{2} = 1.5$

   • $f(1.5) = 2(1.5) - 3 - \cos(1.5) = 3 - 3 - 0.0707 = -0.0707$
   • Since $f(1.5)$ is negative, the root lies between $1.5$ and $2$.

2. Midpoint $c_2 = \frac{1.5 + 2}{2} = 1.75$

   • $f(1.75) = 2(1.75) - 3 - \cos(1.75) = 3.5 - 3 - (-0.1782) = 0.6782$
   • Since $f(1.75)$ is positive, the root lies between $1.5$ and $1.75$.

3. Midpoint $c_3 = \frac{1.5 + 1.75}{2} = 1.625$

   • $f(1.625) = 2(1.625) - 3 - \cos(1.625) = 3.25 - 3 - (-0.0571) = 0.3071$
   • The root lies between $1.5$ and $1.625$.

We continue this process until we reach the desired level of accuracy. After a few more iterations, the root converges to approximately $1.52$.

The positive root is approximately $x = 1.52$.

Would you like a detailed explanation of each iteration, or do you have any questions?

Here are 5 related questions you could explore:

1. What are the advantages and disadvantages of the bisection method?
2. How does the choice of initial interval affect the convergence speed in the bisection method?
3. What is the convergence rate of the bisection method compared to other numerical methods?
4. How would you apply Newton's method to solve $2x = 3 + \cos x$?
5. How can we ensure the accuracy of the root in numerical methods like bisection?

Tip: The bisection method is guaranteed to converge as long as the function is continuous and the initial interval contains a root.