To solve the equation \( \cos 2x - \cos x + 1 = 0 \), we can use trigonometric identities and substitution.

### Step 1: Apply the double-angle identity for cosine
Recall that:
\[
\cos 2x = 2\cos^2 x - 1
\]
Substitute this into the equation:
\[
2\cos^2 x - 1 - \cos x + 1 = 0
\]

### Step 2: Simplify the equation
Combine like terms:
\[
2\cos^2 x - \cos x = 0
\]

### Step 3: Factor the equation
Factor out \( \cos x \):
\[
\cos x (2\cos x - 1) = 0
\]

This gives us two possible cases:
1. \( \cos x = 0 \)
2. \( 2\cos x - 1 = 0 \)

### Step 4: Solve each case

#### Case 1: \( \cos x = 0 \)
If \( \cos x = 0 \), then:
\[
x = \frac{\pi}{2} + n\pi \quad \text{for integer } n
\]

#### Case 2: \( 2\cos x - 1 = 0 \)
Solve for \( \cos x \):
\[
2\cos x = 1
\]
\[
\cos x = \frac{1}{2}
\]
This gives:
\[
x = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{3} + 2n\pi \quad \text{for integer } n
\]

### Final Solution
The solutions to the equation \( \cos 2x - \cos x + 1 = 0 \) are:
\[
x = \frac{\pi}{2} + n\pi, \quad x = \frac{\pi}{3} + 2n\pi, \quad \text{and} \quad x = \frac{5\pi}{3} + 2n\pi
\]
for integer \( n \).

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

Here are some related questions for further exploration:
1. How would this solution change if the equation were \( \cos 2x - 2\cos x + 1 = 0 \)?
2. What are the general solutions for \( \cos^2 x - \cos x = 0 \)?
3. How can we derive the double-angle identities from first principles?
4. How would you solve \( \sin 2x - \sin x = 0 \) using similar techniques?
5. Can you solve \( \cos 3x - \cos x = 0 \) using trigonometric identities?

**Tip:** When solving trigonometric equations, always check if identities can simplify expressions before expanding further.

Learn how to solve the trigonometric equation cos2x - cosx + 1 = 0 by applying the double-angle identity for cosine and factoring. This detailed solution involves solving for the roots of the equation, covering key concepts such as the cosine function and trigonometric identities. Suitable for students in Grades 10-12.

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