Find a Positive Root of 2x = 3 + cos(x) by Bisection method

To find a positive root of the equation \( 2x = 3 + \cos(x) \), we will solve it using the **bisection method**. The bisection method works by iteratively narrowing down an interval where the sign of the function changes, meaning a root exists in that interval.

### Step 1: Rewrite the equation in standard form
First, rewrite the equation as:
\[
f(x) = 2x - 3 - \cos(x) = 0
\]
We need to find a positive value of \( x \) such that \( f(x) = 0 \).

### Step 2: Choose the interval \([a, b]\)
We need to find an interval where \( f(a) \) and \( f(b) \) have opposite signs (indicating that a root lies between \( a \) and \( b \)).

Let’s test a few values for \( f(x) \):

- \( f(0) = 2(0) - 3 - \cos(0) = -3 - 1 = -4 \)
- \( f(1) = 2(1) - 3 - \cos(1) = 2 - 3 - \cos(1) \approx -1.5403 \)
- \( f(2) = 2(2) - 3 - \cos(2) = 4 - 3 - \cos(2) \approx 0.5839 \)

Thus, we have:
\[
f(1) \approx -1.5403 \quad \text{and} \quad f(2) \approx 0.5839
\]
Since \( f(1) \) and \( f(2) \) have opposite signs, there must be a root between \( 1 \) and \( 2 \).

### Step 3: Apply the Bisection Method

Now, we will apply the bisection method, which works by repeatedly halving the interval \([a, b]\) until the root is found with desired accuracy.

#### Iteration 1:
- Midpoint: \( c = \frac{1 + 2}{2} = 1.5 \)
- \( f(1.5) = 2(1.5) - 3 - \cos(1.5) \approx 0.0739 \)

Since \( f(1.5) \approx 0.0739 \) and \( f(1) \approx -1.5403 \), the root lies between \( 1 \) and \( 1.5 \) because they have opposite signs.

#### Iteration 2:
- Midpoint: \( c = \frac{1 + 1.5}{2} = 1.25 \)
- \( f(1.25) = 2(1.25) - 3 - \cos(1.25) \approx -0.7568 \)

Since \( f(1.25) \approx -0.7568 \) and \( f(1.5) \approx 0.0739 \), the root lies between \( 1.25 \) and \( 1.5 \).

#### Iteration 3:
- Midpoint: \( c = \frac{1.25 + 1.5}{2} = 1.375 \)
- \( f(1.375) = 2(1.375) - 3 - \cos(1.375) \approx -0.348 \)

The root is between \( 1.375 \) and \( 1.5 \).

#### Iteration 4:
- Midpoint: \( c = \frac{1.375 + 1.5}{2} = 1.4375 \)
- \( f(1.4375) = 2(1.4375) - 3 - \cos(1.4375) \approx -0.1339 \)

#### Iteration 5:
- Midpoint: \( c = \frac{1.4375 + 1.5}{2} = 1.46875 \)
- \( f(1.46875) = 2(1.46875) - 3 - \cos(1.46875) \approx -0.0300 \)

#### Iteration 6:
- Midpoint: \( c = \frac{1.46875 + 1.5}{2} = 1.484375 \)
- \( f(1.484375) = 2(1.484375) - 3 - \cos(1.484375) \approx 0.0222 \)

#### Iteration 7:
- Midpoint: \( c = \frac{1.46875 + 1.484375}{2} = 1.4765625 \)
- \( f(1.4765625) = 2(1.4765625) - 3 - \cos(1.4765625) \approx -0.0039 \)

Now, the sign changes between \( 1.4765625 \) and \( 1.484375 \). You can continue the iterations until you reach the desired precision.

### Approximate Positive Root
After these iterations, the approximate positive root is around \( x \approx 1.4766 \).

Would you like further details on any step or continue refining the root?

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#### 5 Related Questions:
1. What is the general concept behind the bisection method for solving equations?
2. How does the choice of initial interval affect the performance of the bisection method?
3. Can the bisection method be applied to all types of equations? Why or why not?
4. What are some other numerical methods for finding roots of equations?
5. How do you determine the stopping criteria for the bisection method?

**Tip:** The bisection method works best when the function is continuous and you have an interval where the function changes signs.

Learn how to find a positive root of the equation 2x = 3 + cos(x) using the bisection method. This solution includes detailed steps, key formulas, and application of the Intermediate Value Theorem. Suitable for advanced high school or college students studying numerical methods.

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