Provide a trigonometric equation. Considering only the space between 𝑥 = 0 𝑎𝑛𝑑 2𝜋, the equation must only have solutions at 𝑥 = 1 and 𝑥 = 2. Explain your thought process and the work you did to create the equation. You may round decimal values to 3 places.

To create a trigonometric equation that only has solutions at \( x = 1 \) and \( x = 2 \) within the interval \( 0 \leq x \leq 2\pi \), we need to craft a function that equals zero precisely at those points and nowhere else.

### Step 1: Constructing a Polynomial Form
A common way to ensure that a function has specific zeros is to use a factored polynomial approach. For example, a polynomial of the form:

\[
f(x) = (x - a)(x - b)
\]

has zeros at \( x = a \) and \( x = b \). However, since we are dealing with trigonometric functions, we can use a similar idea but replace \( x \) with trigonometric terms. We'll use cosine or sine, which oscillate and allow us to restrict solutions between \( 0 \) and \( 2\pi \).

### Step 2: Scaling the Domain to Fit the Required Zeros
Since we want solutions at \( x = 1 \) and \( x = 2 \), let's use a shifted cosine function. A cosine function has a range of solutions between 0 and \( 2\pi \), so we will scale and shift it to hit our required points.

To accomplish this, we can use a form like:

\[
\cos(k(x - d))
\]

where \( k \) adjusts the frequency, and \( d \) shifts the function horizontally. We need the function to equal zero at \( x = 1 \) and \( x = 2 \).

### Step 3: Finding the Parameters
The cosine function equals zero at certain points of its cycle. A shifted cosine function, \( \cos(k(x - d)) \), will equal zero when \( k(x - d) \) is an odd multiple of \( \frac{\pi}{2} \). We want it to be zero at \( x = 1 \) and \( x = 2 \), so:

1. Set \( \cos(k(1 - d)) = 0 \) and \( \cos(k(2 - d)) = 0 \).

Since \( \cos(\frac{\pi}{2}) = 0 \), we'll adjust the phase shift and frequency to meet these conditions.

### Step 4: The Final Equation
One equation that satisfies this requirement is:

\[
f(x) = \cos\left( \pi(x - 1.5) \right)
\]

This function equals zero at \( x = 1 \) and \( x = 2 \), and these are the only solutions between \( 0 \) and \( 2\pi \).

### Thought Process Summary:
1. We aimed to create a trigonometric function that has zeros only at \( x = 1 \) and \( x = 2 \).
2. We used the cosine function, which naturally oscillates and can be shifted to achieve the desired zeros.
3. We scaled the function so that it equals zero precisely at the required points and adjusted it to fit within the interval \( [0, 2\pi] \).

Would you like more details on any step, or do you have further questions?

---

Here are five related questions:
1. How does shifting a cosine function affect its graph?
2. Why does cosine have specific zeros at \( \frac{\pi}{2} \), \( \frac{3\pi}{2} \), etc.?
3. Can a sine function be used to achieve the same solutions?
4. How can you modify the frequency of a trigonometric function to alter its zeros?
5. What is the general method for finding zeros of a trigonometric equation?

**Tip:** Always start by identifying key points (like zeros or peaks) and use shifts and scaling to manipulate trigonometric functions effectively.

Provide a trigonometric equation. Considering only the space between 𝑥 = 0 and 2𝜋, the equation must only have solutions at 𝑥 = 1 and 𝑥 = 2. Explain your thought process and the work you did to create the equation. You may round decimal values to 3 places.

Learn how to create a trigonometric equation with solutions at x = 1 and x = 2 between 0 and 2π. This problem explores cosine function shifts, zeros, and phase adjustments, suitable for students in Grades 10-12.

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